Let $X', Y', P'$, and $Q'$ be the reflections across $BC$ of $X, Y, P$, and $Q$, respectively. Then $\Omega_1$ and $\Omega_2$ are just the circles $PXX'P'$ and $QYY'Q'$, respectively.
Denote $\alpha = \angle BAC = \angle CBX = \angle YCB$. Let $XY$ cross $BC$ at $W$; the case $XY \parallel BC$ may be treated as a limit case. The symmetry yields that $W$ also lies on the line $X'Y'$. The same symmetry, along with tangency of $PB$ and $QC$ to the circle $ABC$, yields
$$\alpha = \angle X'BC = \angle PBW = \angle WCQ = \angle BCY'. \quad (*)$$
This yields that each of the triples $(P, B, X')$, $(Q, C, Y')$, $(P', B, X)$, and $(Q', C, Y)$ is collinear, and, moreover, that $PBX' \parallel Q'CY$ and $P'BX \parallel QCY'$. It follows now that quadrilaterals $PXX'P'$ and $YQQ'Y'$ are homothetic at $W$. Therefore, so are $\Omega_1$ and $\Omega_2$.

Let now $\Omega_1$ and $\Omega_2$ cross at $D$ and $D'$. Let $WD$ and $WD'$ meet $\Omega_1$ again at $E$ and $E'$. Since $W = PX \cap P'X'$ and $B = PX' \cap P'X$, the point $B$ lies on the polar of $W$ with respect to $\Omega_1$. In other words, $W$ and $B$ are inverse with respect to that circle. This yields that the lines $DE'$ and $D'E$ also cross at $B$.

Now, we have $\angle BDX = \angle E'DX = \angle X'D'E = \angle X'PE = \angle BPE = \angle CYD$ (the last equality holds by means of homothety). Similarly, we have $\angle DXB = \angle DXP' = \angle PX'D' = \angle PED' = \angle PEB = \angle YDC$. Therefore, the triangles $BDX$ and $CYD$ are similar. Firstly, this yields that $\angle DBC = \angle DBX + \angle XBC = \angle YCD + \angle BCY = \angle BCD$, whence $BD = CD$. Secondly, this also implies that $BD/BX = CY/CD$, or $BX \cdot CY = BD \cdot CD = BD^2$. But the triangles $BXC$ and $CBY$ are also similar (as both are similar to $ABC$), so $BX/BC = BC/CY$, or $BX \cdot CY = BC^2$. Thus, $BC = BD = CD$, and the triangle $BCD$ is equilateral. This finishes the solution.

Solution 2:
All angles in the solution are directed. All segment lengths on lines $BX$ and $CY$ (and parallel to them) are also oriented; we assume that the directions $\overrightarrow{BX}$ and $\overrightarrow{CY}$ are positive. As in the solution above, we prove that $BP \parallel CY$.
Assume that $ABC$ is oriented anti-clockwise. Let $D$ and $D'$ be the points such that the triangles $DBC$ and $D'CB$ are equilateral, and oriented anti-clockwise. We will show that $D$ and $D'$ lie on the circle $\Omega_1$; similarly, they lie on $\Omega_2$.
Notice that $\alpha = \angle BAC = \angle CBX = \angle YCB = \pi - \angle CBP$; moreover, each of the triangles $XBC$ and $BCY$ is similar to $BAC$ and oriented differently than $BAC$; hence those two triangles are equi-oriented. Let $\Omega$ denote the circle $(DD'X)$; clearly, its centre lies on the perpendicular bisector of $DD'$, i.e., on $BC$. We aim to prove that $\Omega$ passes through $P$; that will yield that $\Omega = \Omega_1$, which establishes what we are aimed to prove.
Denote $Z = XB \cap YC$. Since $\angle CBZ = \angle BAC = \angle ZCB$, we have $ZB = ZC$, and hence $Z$ lies on the perpendicular bisector $DD'$ of $BC$. By similarity, we get $BX/BC = BC/CY$, or $BC^2 = BX \cdot CY = BX \cdot (ZY + CZ)$. Since $CY \parallel BP$, the triangles $XZY$ and $XBP$ are similar, so $BX \cdot ZY = ZX \cdot BP$.

Therefore, $BD^2 = BC^2 = BX \cdot ZY + BX \cdot CZ = ZX \cdot BP + (BZ + ZX) \cdot BZ = ZX \cdot (BP + BZ) + BZ^2$.
On the other hand, let $M$ be the midpoint of $BC$, and let $XB$ cross $\Omega$ again at $P'$. Write the power of point $Z$ with respect to $\Omega$ as $XZ \cdot (P'B + BZ) = XZ \cdot P'Z = ZD \cdot ZD' = MZ^2 - DM^2 = BZ^2 - MB^2 - DM^2 = BZ^2 - BD^2$.
The two obtained relations yield
$$ZX \cdot (BP + BZ) = BD^2 - BZ^2 = ZX \cdot (P'B + BZ),$$
so $BP = P'B$, and so $P$ and $P'$ are reflections of one another in the line $BC$. Thus, $P$ lies on $\Omega$, as desired.

Remarks.
(1) It is also possible to solve the problem via the moving points method. Introduce the points $D$ and $D'$ as in Solution 2, and introduce the reflections $X'$, $Y'$, $P'$, and $Q'$ of $X$, $Y$, $P$, and $Q$ in the line $BC$, respectively, as in Solution 1, to read $\Omega_1$ and $\Omega_2$ as the circles $PXX'P'$ and $QYY'Q'$, respectively.
We need to show that $D$ lies on $\Omega_2$ (the other incidences are similar). To this end, it suffices to check that $\angle YDQ = \angle YY'Q = 90^\circ - \angle Y'CB = 90^\circ - \angle BAC$.
Fix $B$, $C$, and the circle $ABC$. As $A$ varies on that circle, the lines $BX$, $CY$, $BP$, and $CQ$ remain constant, and $X$ and $Y$ depend projectively on $A$. Choosing $Q_1$ on $CQ$ such that $\angle YDQ_1 = 90^\circ - \angle BAC$, we need to show that $Q_1 = Q$, or that $X$, $Y$, and $Q_1$ are collinear. The point $Q_1$ also depends projectively on $A$, so it suffices to check that the points $Q_1$, $X$, and $Y$ are collinear for four specific positions of $A$.

(2) Although inversion, homothety, and the moving point method are essentially the same thing, i.e., a Möbius transformation, we briefly sketch yet another approach below:
Let $BP \cap CQ = D$, $BX \cap CY = Z$, and $DZ \cap XY = T$. Let $R_1$ and $R_2$ be distinct points such that triangles $R_1BC$ and $R_2BC$ are equilateral. We first note that $BZ \parallel CD$ and $CZ \parallel BD$, so the triangles $DPQ$ and $ZYX$ are homothetic from $T$. Hence, $TP \cdot TX = TQ \cdot TY$, so $T$ lies on the radical axis of $\Omega_1$ and $\Omega_2$. Since their centres both lie on $BC$, their radical axis is the perpendicular from $T$ to $BC$, i.e. the perpendicular bisector $DZ$ of $BC$. As $CY \parallel DP$ and $BX \parallel DQ$, it follows that $BP = YZ \cdot \frac{BX}{XZ}$ and $CQ = XZ \cdot \frac{CY}{YZ}$. As triangles $BXC$ and $CYB$ are similar, this means that $BP \cdot CQ = BX \cdot CY = BC^2$.
Perform an inversion about $\Omega$, the circle of radius $BC$ centred at $B$. Let $X'$, $P'$, and $\Omega'_1$ denote the images of $X$, $P$ and $\Omega_1$. As $BX \cdot CY = BP \cdot CQ = BC^2$, we get $BP' = CQ$ and $BX' = CY$. Since lines $BD$, $CD$ and $BZ$, $CZ$ are symmetric about $DZ$, it follows that $X'$, $P'$ and $Y$, $Q$ are symmetric about $DZ$. Thus, $\Omega'_1$ and $\Omega_2$ are reflections about $DZ$. Since $\Omega_1 \cap \Omega_2$ and $\Omega'_1 \cap \Omega_2$ all lie on $DZ$, it follows that $\Omega$, $\Omega_1$, $\Omega'_1$, $\Omega_2$ all meet on $DZ$. This implies that $\Omega_1$ and $\Omega_2$ both pass through $R_1$ and $R_2$, the two required fixed points.