Denote by $\omega$ the circumcircle of the triangle $ABC$, and let $\angle ACB = \gamma$. Note that the condition $\gamma < \angle CBA$ implies $\gamma < 90^\circ$. Since $\angle PBA = \gamma$, the line $PB$ is tangent to $\omega$, so $PA \cdot PC = PB^2 = PD^2$. By $\frac{PA}{PD} = \frac{PD}{PC}$ the triangles $PAD$ and $PDC$ are similar, and $\angle ADP = \angle DCP$.

Next, since $\angle ABQ = \angle ACB$, the triangles $ABC$ and $AQB$ are also similar. Then $\angle AQB = \angle ABC = \angle ARC$, which means that the points $D, R, C$, and $Q$ are concyclic. Therefore $\angle DRQ = \angle DCQ = \angle ADP$.

Figure 1

Now from $\angle ARB = \angle ACB = \gamma$ and $\angle PDB = \angle PBD = 2\gamma$ we get
$$\angle QBR = \angle ADB - \angle ARB = \angle ADP + \angle PDB - \angle ARB = \angle DRQ + \gamma = \angle QRB,$$
so the triangle $QRB$ is isosceles, which yields $QB = QR$.

Again, denote by $\omega$ the circumcircle of the triangle $ABC$. Denote $\angle ACB = \gamma$. Since $\angle PBA = \gamma$, the line $PB$ is tangent to $\omega$.

Let $E$ be the second intersection point of $BQ$ with $\omega$. If $V'$ is any point on the ray $CE$ beyond $E$, then $\angle BEV' = 180^\circ - \angle BEC = 180^\circ - \angle BAC = \angle PAB$; together with $\angle ABQ = \angle PBA$ this shows firstly, that the rays $BA$ and $CE$ intersect at some point $V$, and secondly that the triangle $VEB$ is similar to the triangle $PAB$. Thus we have $\angle BVE = \angle BPA$. Next, $\angle AEV = \angle BEV - \gamma = \angle PAB - \angle ABQ = \angle AQB$; so the triangles $PBQ$ and $VAE$ are also similar.

Let $PH$ be an altitude in the isosceles triangle $PBD$; then $BH = HD$. Let $G$ be the intersection point of $PH$ and $AB$. By the symmetry with respect to $PH$, we have $\angle BDG = \angle DBG = \gamma = \angle BEA$; thus $DG \parallel AE$ and hence $\frac{BG}{GA} = \frac{BD}{DE}$. Thus the points $G$ and $D$ correspond to each other in the similar triangles $PAB$ and $VEB$, so $\angle DVB = \angle GPB = 90^\circ - \angle PBQ = 90^\circ - \angle VAE$. Thus $VD \perp AE$.

Let $T$ be the common point of $VD$ and $AE$, and let $DS$ be an altitude in the triangle $BDR$. The points $S$ and $T$ are the feet of corresponding altitudes in the similar triangles $ADE$ and $BDR$, so $\frac{BS}{SR} = \frac{AT}{TE}$. On the other hand, the points $T$ and $H$ are feet of corresponding altitudes in the similar triangles $VAE$ and $PBQ$, so $\frac{AT}{TE} = \frac{BH}{HQ}$. Thus $\frac{BS}{SR} = \frac{AT}{TE} = \frac{BH}{HQ}$, and the triangles $BHS$ and $BQR$ are similar.

Finally, $SH$ is a median in the right-angled triangle $SBD$; so $BH = HS$, and hence $BQ = QR$.

Figure 2

Denote by $\omega$ and $O$ the circumcircle of the triangle $ABC$ and its center, respectively. From the condition $\angle PBA = \angle BCA$ we know that $BP$ is tangent to $\omega$.

Let $E$ be the second point of intersection of $\omega$ and $BD$. Due to the isosceles triangle $BDP$, the tangent of $\omega$ at $E$ is parallel to $DP$ and consequently it intersects $BP$ at some point $L$. Of course, $PD \parallel LE$. Let $M$ be the midpoint of $BE$, and let $H$ be the midpoint of $BR$. Notice that $\angle AEB = \angle ACB = \angle ABQ = \angle ABE$, so $A$ lies on the perpendicular bisector of $BE$; thus the points $L, A, M$, and $O$ are collinear. Let $\omega_1$ be the circle with diameter $BO$. Let $Q' = HO \cap BE$; since $HO$ is the perpendicular bisector of $BR$, the statement of the problem is equivalent to $Q' = Q$.

Consider the following sequence of projections (see Fig. 3).
1. Project the line $BE$ to the line $LB$ through the center $A$. (This maps $Q$ to $P$.)
2. Project the line $LB$ to $BE$ in parallel direction with $LE$. ($P \mapsto D$.)
3. Project the line $BE$ to the circle $\omega$ through its point $A$. ($D \mapsto R$.)
4. Scale $\omega$ by the ratio $\frac{1}{2}$ from the point $B$ to the circle $\omega_1$. ($R \mapsto H$.)
5. Project $\omega_1$ to the line $BE$ through its point $O$. ($H \mapsto Q'$.)

We prove that the composition of these transforms, which maps the line $BE$ to itself, is the identity. To achieve this, it suffices to show three fixed points. An obvious fixed point is $B$ which is fixed by all the transformations above. Another fixed point is $M$, its path being $M \mapsto L \mapsto E \mapsto E \mapsto M \mapsto M$.

Figure 3

Figure 4

In order to show a third fixed point, draw a line parallel with $LE$ through $A$; let that line intersect $BE, LB$ and $\omega$ at $X, Y$ and $Z \neq A$, respectively (see Fig. 4). We show that $X$ is a fixed point. The images of $X$ at the first three transformations are $X \mapsto Y \mapsto X \mapsto Z$. From $\angle XBZ = \angle EAZ = \angle AEL = \angle LBA = \angle BZX$ we can see that the triangle $XBZ$ is isosceles. Let $U$ be the midpoint of $BZ$; then the last two transformations do $Z \mapsto U \mapsto X$, and the point $X$ is fixed.