The center of $\omega$ is $D$, the point of $\Gamma$ diametrically opposed to $A$. Let $O$ be the center of $\Gamma$. Denote by $\lambda=\angle BCP$ and $\mu=\angle PBC$. It is clear that $\lambda+\mu$ is constant. The angles $\lambda,\mu$ appear again many times, e.g. $\angle ABP=\lambda$, $\angle BAR=\angle BCR=\lambda$, $\angle PCA=\mu$, $\angle RBA=\angle RCA=\mu$. Now, because triangles $ABX$ and $ACY$ are isosceles we have $\angle YAC=\lambda$ and $\angle BAX=\mu$. Adding known angles, it is clear that angle $\angle YAX$ is constant: $$ \angle YAX=\angle YAC+\angle CAB+\angle BAX=\lambda + 180^o-2\lambda-2\mu + \mu=180^o-\lambda-\mu. $$We conjecture that the second fixed point of the circumcircle of $AXY$ is $D$. To prove this, we need that $\angle XDY=\lambda+\mu$. Consider the point $Z$, the reflection of $Y$ with respect to $AY$. Since $\angle ZAB=\angle YAC=\lambda$, it follows that $Z$ lies on the line $AR$. Also, $Z$ belongs to $OX$. In fact, $Z$ is the inverse of $X$ in the circle $\Gamma$, because triangles $OAZ$ and $OXA$ are inversely similar (with angles $\lambda+\mu, 90^o-\mu, 90^o-\lambda$, easily obtained by adding and subtracting already known angles). From this, we have $OX\cdot OZ=OA^2=OD^2$. This means that triangles $OXD$ and $ODZ$ are inversely similar. But $ODZ$ and $ODY$ are inversely similar, from which it follows that $OXD$ and $ODY$ are directly similar. Finally: $$ \angle XDY=\angle XDO+\angle ODY=\angle XDO+\angle OXD=180^o-\angle DOX=\lambda+\mu. $$

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Denote $(X) = (X, XA), (Y) = (Y, YA)$. Let $F \neq A$ be the second intersection point of $(X)$ and $(Y)$. $\angle{BFC} = 360^{\circ} - \angle{BFA} - \angle{CFA} = \frac{\angle{BXA} + \angle{CYA}}{2} = \frac{360^{\circ} - 2\angle{RBA} - 2\angle{QCA}}{2} = 180^{\circ} - \angle{PBC} - \angle{PCB} = \angle{BPC}$, thus $F$ lies on $\omega$. Let $I$ be the center of $\omega$. Since the radical axis of two circles is perpendicular to the line passing through their centers, we have $XI \perp BF$ and $YI \perp CF$. Therefore, $\angle{XIY} = 180^{\circ} - \angle{BFC}$. Simple angle chasing gives us that $\angle{XAY} = \angle{BPC} = \angle{BFC}$, thus $(XAY)$ passes through $I$ which is fixed while $P$ is moving.

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Let $O$ be the circumcenter of the circle $(ABC)$, and let the midpoints of $AB$ and $AC$ be $M$ and $N$ respectively. Denote by $T$ the arc midpoint of $BC$ not containing $A$, i.e. the center of $\omega$. $\textbf{Claim.} OX \cdot OY = R^2,$ where $R$ is the circumradius. $\textit{Proof.}$ Let the reflection of $Y$ over $AO$ be $Y'$. It is enough to show that $X$ and $Y'$ are inverses with respect to $\Gamma$. Now observe that \[ \measuredangle AQB = \measuredangle ACB = \measuredangle CPB = \measuredangle RPQ, \]so $AQ \parallel PR$. For analogous reasons we obtain that $AQPR$ is a parallelogram. In particular we obtain the implication that $BQAR$ is an isosceles trapezoid. Let $Q'$ be the reflection of $Q$ over $AO$. Observe that $BR=AQ=AQ'$ so we have $RQ \parallel AB$. However this means that the angle bisector of $RBQ'$ is fixed, and in particular it passes through the arc midpoint of $AB$ not containing $C$. Let this point be $U$, and let the antipodal point of $U$ w.r.t. $\Gamma$ be $V$. By the so-called "Right angles and bisectors" lemma, we obtain that $(X, Y'; U, V) = -1$; however this is enough to imply that $X$ and $Y'$ are inverses because $UV$ is a diameter of $\Gamma$. Therefore $XO \cdot Y'O = R^2$. $\square$ Returning, if we let $XO$ intersect the circle $(AXT)$ at point $Z$, we obtain that $XO \cdot OZ = AO \cdot OT = R^2$, so $OZ = OY$. Moreover this implies that $Z$ and $Y$ are reflections over the perpendicular bisector of $AT$, so $Y$ lies on $(AXT)$. Therefore the circumcircle of $(AXY)$ passes through $T$.