Solution:

Assume that $\omega$ is tangent to $A C$ and $B C$ at $E$ and $F$, respectively and let $P E, P F, P Q$ meet $\Omega$ at $K, L, M$, respectively. Let $I$ and $O$ denote the respective centres of $\omega$ and $\Omega$, and consider the homothety $\mathscr{H}$ that maps $\omega$ onto $\Omega$. Now $K$ is the image of $E$ under $\mathscr{H}$, and $E I \perp A C$. Hence $O K \perp A C$, and thus $K$ is the midpoint of the arc $C A$. Similarly, $L$ is the midpoint of the arc $B C$ and $M$ is the midpoint of the arc $B A$. It follows that arcs $L M$ and $C K$ are equal, because
$$\text{B M = M A B L + L M = M K + K A L C + L M = M K + C K 2 L M + M C = M C +2 C K}$$
Thus arcs $F Q$ and $D E$ of $\omega$ are equal, too, where $D$ is the intersection of $C P$ with $\omega$. Since $C E$ and $C F$ are tangents to $\omega$, this implies that $\angle D E C=\angle C F Q$. Further, $C E=C F$, and thus triangles $C E D$ and $C F Q$ are congruent. In particular, $\angle E C D=\angle Q C F$, as required.

A Variant. As above, we show that $\operatorname{arcs} F Q$ and $D E$ of $\omega$ are equal, which implies that $D E F Q$ is an isoceles trapezoid, and so we have $\angle F E D=\angle Q F E$. Together with $|F Q|=|D E|$, this implies that, since $E$ and $F$ are images of each other under reflection in the angle bisector $C I$ of $\angle C$, so are the segments $[E Q]$ and $[F D]$, and, in particular, $D$ and $Q$. In turn, this yields $\angle E C D=\angle Q C F$, as required.

Remark. Let $J$ denote the incentre of $A B C$. By Sawayama's theorem, $J$ is the midpoint of $[E F]$, i.e. $P J$ is a median of $P F E$. Since $C$ is the intersection of the tangents $A C$ and $B C$ to the circumcircle of $P F E$ at $E$ and $F$, respectively, $P C$ is a symmedian of $P F E$. Thus $\angle C P E=\angle F P J$. But, since the arcs $F Q$ and $D E$ of $\omega$ are equal, $\angle C P E=\angle F P Q$. This shows that $J$ lies on the line $P Q$.

Another Variant. We show that $\operatorname{arcs} Q E$ and $F D$ are equal, and then finish as in the main solution. Let $B P$ meet $\omega$ again at $Z$. Consider the homothety $\mathscr{H}$ that maps $\omega$ onto $\Omega$. Under $\mathscr{H}, D \mapsto C$ and $Z \mapsto B$, so $D Z \parallel C B$. (This also follows by considering the common tangent to $\omega$ and $\Omega$, and tangential angles.) Now, by power of a point,
$$B F^{2}=B Z \cdot B P, \quad C F^{2}=C D \cdot C P$$
Now $D Z \parallel C B$ implies $B Z / B P=C D / C P$, and so, dividing the two previous equations by each other, and taking square roots, $B F / C F=B P / C P$. Hence $P F$ bissects angle $\angle B P C$. Now let $\angle B P F=\angle F P C=\beta$. By tangential angles, it follows that $\angle C F D=\beta$. Further, $\angle B A C=\angle B P C=2 \beta$. Let the tangent to $\omega$ through $Q$ and parallel to $A B$ meet $A C$ at $X$. Then $\angle Q X C=2 \beta$, so, since $X Q=X E$ by tangency, $\angle Q E X=\beta$. By tangential angles, it follows that $\operatorname{arcs} F D$ and $Q E$ are equal, as claimed.

Solution:

Let $I$ and $O$ denote the respective centres of $\omega$ and $\Omega$. Observe that $C I$ is the angle bisector of angle $\angle C$, because $\omega$ is tangent to $A C$ and $B C$. Consider the homothety $\mathscr{H}$ that maps $\omega$ onto $\Omega$. Let $M$ be the image of $Q$ under $\mathscr{H}$. By construction, $I Q \perp A B$, so $O M \perp A B$. Thus the diameter $O M$ of $\Omega$ passes through the midpoint of the $\operatorname{arc} A B$ of $\Omega$, which also lies on the angle bisector $C I$. This implies that $\angle I C M=90^{\circ}$. We next show that $P, I, Q, C$ lie on a circle. Notice that
$$\begin{aligned} \angle P Q I & =90^{\circ}-\frac{1}{2} \angle Q I P=90^{\circ}-\frac{1}{2} \angle M O P=90^{\circ}-\left(180^{\circ}-\angle P C M\right) \\ & =(\angle P C I+\angle I C M)-90^{\circ}=\angle P C I \end{aligned}$$
Hence $P, I, Q, C$ lie on a circle. But $P I=I Q$, so $C I$ is the angle bisector of $\angle P C Q$. Since $C I$ is also the angle bisector of angle $\angle C$, it follows that $\angle A C P=\angle Q C B$, as required.

A Variant. We show that $P I Q C$ is cyclic by chasing angles. Define $\alpha=\angle B A C$, $\beta=\angle C B A$ and $\gamma=\angle A C P$. For convenience, we consider the configuration where $A$ and $P$ lie one the same side of the angle bisector $C I$ of $\angle C$. In this configuration,
$$\angle P C I=\frac{1}{2} \angle A C B-\angle A C P=90^{\circ}-\frac{1}{2} \alpha-\frac{1}{2} \beta-\gamma .$$
Now notice that $\angle P B A=\angle A C P=\gamma$, and therefore $\angle C A P=180^{\circ}-\beta-\gamma$, whence $\angle P A B=180^{\circ}-\alpha-\beta-\gamma$. Further, $P O$ is a diameter of $\Omega$, and therefore $\angle A P O=90^{\circ}-\gamma$. Let $A B$ and $P O$ intersect at $T$. Then
$$\angle B T O=180^{\circ}-\angle P A B-\angle A P O=\alpha+\beta+2 \gamma-90^{\circ}$$
But $Q I \perp A B$ by construction, and thus
$$\begin{aligned} & \angle O I Q=90^{\circ}-\angle B T O=180^{\circ}-\alpha-\beta-2 \gamma \\ & \Longrightarrow \quad \angle Q I P=180^{\circ}-\angle O I Q=\alpha+\beta+2 \gamma \\ & \Longrightarrow \quad \angle P Q I=90^{\circ}-\frac{1}{2} \alpha-\frac{1}{2} \beta-\gamma . \end{aligned}$$
Hence $\angle I C Q=\angle P Q I$, and thus $P I Q C$ is cyclic. Since $P I=Q I$, it follows that $C I$ is the angle bisector of $\angle P C Q$, which completes the proof.

Solution:

Let $I$ and $O$ denote the respective centres of $\omega$ and $\Omega$. Let $D$ be the second point of intersection of $C P$ with $\omega$, and let $\ell$ denote the tangent to $\omega$ at $D$, which meets $A C$ at $S$. Hence $I D \perp \ell$. By construction, $P, I, O$ lie one a line, and hence the isosceles triangles $P I D$ and $P O C$ are similar. In particular, it follows that $O C \perp \ell$, so $C$ is the midpoint of the arc of $\Omega$ defined by the points of intersection of $\ell$ with $\Omega$. It is easy to see that this implies that
$$\angle D S C=\angle A B C$$
Under reflection in the angle bisector $C I$ of $\angle C, \ell$ is thus mapped to a tangent to $\omega$ parallel to $A B$ and intersecting the interior of $A B C$, since $\omega$ is mapped to itself under this reflection. In particular, $D$ is mapped to $Q$, and thus $\angle Q C B=\angle A C D$, as required.

Remark. Conceptually, this solution is similar to Solution 1, but here, we proceed more directly via the reflectional symmetry. Therefore, this solution links Solution 1 to Solution 4, in which we use an inversion.

Solution:

Let the tangent to $\omega$ at $Q$ meet $A C$ and $B C$ at $X$ and $Y$, respectively. Then $A C / X C=B C / Y C$, and thus there is a radius $r$ such that $r^{2}=A C \cdot Y C=B C \cdot X C$. Let $\Gamma$ denote the circle with centre $C$ and radius $r$, and consider the inversion $\mathscr{I}$ in the circle $\Gamma$. Under $\mathscr{I}$,
$A \longmapsto A^{\prime}$, the point on the ray $C A$ satisfying $C A^{\prime}=C Y$;
$B \longmapsto B^{\prime}$, the point on the ray $C B$ satisfying $C B^{\prime}=C X$;
$\Omega \longmapsto$ the line $A^{\prime} B^{\prime}$;
$\omega \longmapsto \omega^{\prime}$, the excircle of $C A^{\prime} B^{\prime}$ opposite $C$;
$P \longmapsto P^{\prime}$, the point where $\omega^{\prime}$ touches $A^{\prime} B^{\prime}$;
In particular, $\omega^{\prime}$, the image of $\omega$, is a circle tangent to $A C, B C$ and $A^{\prime} B^{\prime}$, so it is either the excircle of $C A^{\prime} B^{\prime}$ opposite $C$, or the incircle of $C A^{\prime} B^{\prime}$. Let $\omega$ be tangent to $B C$ at $F$, and let $F^{\prime}$ be the image of $F$ under $\mathscr{I}$. Then $C F \cdot C F^{\prime}=B C \cdot X C$. Now $C F<B C$, so $C F^{\prime}>C X=C B^{\prime}$. Hence $\omega^{\prime}$ cannot be the incircle, so $\omega^{\prime}$ is indeed the excircle of $C A^{\prime} B^{\prime}$ opposite $C$.
Now note that $\omega$ is the excircle of $C X Y$ opposite $C$. The reflection about the angle bisector of $\angle C$ maps $X$ to $B^{\prime}, Y$ to $A^{\prime}$. It thus maps the triangle $C X Y$ to $C B^{\prime} A^{\prime}, \omega$ to $\omega^{\prime}$ and, finally, $Q$ to $P^{\prime}$. It follows that $\angle A C P=\angle A C P^{\prime}=\angle Q C B$, as required.

Solution:

Let $r$ be the radius such that $r^{2}=A C \cdot B C$. Let $\mathscr{J}$ denote the composition of the inversion $\mathscr{I}$ in the circle of centre $C$ and radius $r$, followed by the reflection in the angle bisector of $\angle C$. Under $\mathscr{J}$,
$A \longmapsto B, B \mapsto A ;$
$\Omega \longmapsto$ the line $A B$;
$\omega \longmapsto \omega^{\prime}$, the excircle of $A B C$ opposite the vertex $C$;
$P \longmapsto Q^{\prime}$, the point where $\omega^{\prime}$ touches $A B$;
In particular, note that the image $\omega^{\prime}$ of $\omega$ under $\mathscr{J}$ is a circle tangent to $A C, B C$ and $A B$, so it is either the incircle of $A B C$, or the excircle opposite vertex $C$. Observe that $r \geqslant \min \{A C, B C\}$, so the image of the points of tangency of $\omega$ must lie outside $A B C$, and thus $\omega^{\prime}$ cannot be the incircle. Thus $\omega^{\prime}$ is the excircle opposite vertex $C$ as claimed. Further, the point of tangency $P$ is mapped to $Q^{\prime}$.
Now, since the line $C P$ is mapped to itself under the inversion $\mathscr{I}$, and mapped onto $C Q^{\prime}$ under $\mathscr{J}, C P$ and $C Q^{\prime}$ are images of each other under reflection in the angle bisector of $\angle C$. But $C, Q, Q^{\prime}$ lie on a line for there is a homothety with centre $C$ that maps $\omega$ onto the excircle $\omega^{\prime}$. This completes the proof.

Solution:

Assume that $\omega$ is tangent to $A C$ and $B C$ at $E$ and $F$, respectively. Assume that $C P$ meets $\omega$ at $D$. Let $I$ and $O$ denote the respective centres of $\omega$ and $\Omega$. To set up a solution in the complex plane, we take the circle $\omega$ as the unit circle centered at the origin of the complex plane, and let $P O$ be the real axis with $o>0$, where we use the convention that lowercase letters denote complex coordinates of corresponding points in the plane denoted by uppercase letters.
Now, a point $Z$ on the circle $\Omega$ satisfies
$$|z-o|^{2}=(o+1)^{2} \quad \Longleftrightarrow \quad z z^{*}-o\left(z+z^{*}\right)-2 o-1=0$$
The triangle $A B C$ is defined by the points $E$ and $F$ on $\omega$, the intersection $C$ of the corresponding tangents lying on $\Omega$. Thus $c=2 e f /(e+f)$, and further
$$|c-o|^{2}=(o+1)^{2} \quad \Longleftrightarrow \quad c c^{*}-o\left(c+c^{*}\right)-2 o-1=0,$$
and this is the equality defining $o$. The points $A$ and $B$ are the second intersection points of $\Omega$ with the tangents to $\omega$ at $E$ and $F$ respectively. A point $Z$ on the tangent through $E$ is given by $z=2 e-e^{2} z^{*}$, and thus $A$ and $C$ satisfy
$$\begin{aligned} \left(2 e-e^{2} z^{*}\right) z^{*} & -o\left(2 e-e^{2} z^{*}+z^{*}\right)-2 o-1=0 \\ & \Longleftrightarrow-e^{2} z^{* 2}+\left(2 e+o e^{2}-o\right) z^{*}-(2 e o+2 o+1)=0 \\ & \Longleftrightarrow z^{* 2}-\left(2 e^{*}+o-o e^{* 2}\right) z^{*}+\left(2 e^{*} o+2 o e^{* 2}+e^{* 2}\right)=0 \end{aligned}$$
since $|e|=1$. Thus
$$a^{*}+c^{*}=2 e^{*}+o-o e^{* 2} \quad \Longrightarrow \quad a^{*}=\frac{2 e^{*} f}{e+f}+o\left(1-e^{* 2}\right)$$
and similarly
$$b^{*}=\frac{2 f^{*} e}{f+e}+o\left(1-f^{* 2}\right)$$
Then
$$\begin{aligned} b^{*}-a^{*} & =\frac{2\left(e f^{*}-e^{*} f\right)}{e+f}+o\left(e^{* 2}-f^{* 2}\right) \\ & =\frac{2 e f\left(f^{* 2}-e^{* 2}\right)}{e+f}+o\left(e^{* 2}-f^{* 2}\right) \\ & =\left(f^{* 2}-e^{* 2}\right)\left(\frac{2 e f}{e+f}-o\right)=\left(f^{* 2}-e^{* 2}\right)(c-o) . \end{aligned}$$
Now let $Z$ be a point on the tangent to $\omega$ parallel to $A B$ passing through $Q$. Then
$$z=2 q-q^{2} z^{*} \quad \Longleftrightarrow \quad z-q=q-q^{2} z^{*}=-q^{2}\left(z^{*}-q^{*}\right)$$
for $|q|=1$, and thus
$$\frac{b-a}{b^{*}-a^{*}}=\frac{z-q}{z^{*}-q^{*}}=\frac{-q^{2}\left(z^{*}-q^{*}\right)}{z^{*}-q^{*}}=-q^{2} .$$
It follows that
$$\begin{aligned} q^{2} & =-\frac{b-a}{b^{*}-a^{*}}=-\frac{\left(f^{2}-e^{2}\right)\left(c^{*}-o\right)}{\left(f^{* 2}-e^{* 2}\right)(c-o)}=e^{2} f^{2} \frac{c^{*}-o}{c-o} \\ & =e^{2} f^{2} \frac{\left(c^{*}-o\right)^{2}}{|c-o|^{2}}=e^{2} f^{2} \frac{\left(c^{*}-o\right)^{2}}{(1+o)^{2}} \end{aligned}$$
where we have used (1). In particular,
$$q=e f \frac{c^{*}-o}{1+o}$$
where the choice of sign is to be justified a posteriori. Further, the point $D$ satisfies
$$-d p=\frac{d-p}{d^{*}-p^{*}}=\frac{c-p}{c^{*}-p^{*}} \quad \Longrightarrow \quad d=-\frac{c-p}{c^{*} p-1}=\frac{c+1}{c^{*}+1},$$
using $p=-1$ to obtain the final equality.
Now, it suffices to show that (i) $D Q \parallel E F \perp C I$ and (ii) the midpoint of $[D Q]$ is on $C I$. The desired equality then follows by symmetry with respect to the angle bisector of the angle $\angle A C B$. Notice that (i) is equivalent with
$$\frac{d-q}{d^{*}-q^{*}}=\frac{e-f}{e^{*}-f^{*}} \quad \Longleftrightarrow \quad d q=e f$$
for $[D Q]$ and $[E F]$ are chords of $\omega$. But
$$\begin{aligned} d q=e f & \Longleftrightarrow \frac{c+1}{c^{*}+1} e f \frac{c^{*}-o}{1+o}=e f \quad \Longleftrightarrow \quad(c+1)\left(c^{*}-o\right)=\left(c^{*}+1\right)(1+o) \\ & \Longleftrightarrow c c^{*}-o\left(c+c^{*}\right)-2 o-1=0 . \end{aligned}$$
The last equality is precisely the defining relation for $o,(1)$. This proves (i). Further, the midpoint of $[D Q]$ is $\frac{1}{2}(d+q)$, so it remains to check that
$$d q=\frac{d+q}{d^{*}+q^{*}}=\frac{c}{c^{*}}=e f$$
where the first equality expresses that $[D Q]$ is a chord of $\omega$ (obviously) containing its midpoint, the second equality expresses the alignment of the midpoint of $[D Q], C$ and $I$, and the third equality follows from the expression for $c$. But we have just shown that $d q=e f$. This proves (ii), justifies the choice of sign for $q$ a posteriori, and thus completes the solution of the problem.