Let $C'$ be the $C$-antipode of $\triangle ABC$, $H$ be the orthocenter of $\triangle ABC$ and $M$ be the midpoint of arc $BC$ not containing $A$. Notice that $\triangle AC'M \sim\triangle XHA$ and $\triangle BC'M\sim\triangle XAP$ therefore $$\frac{AP}{MC'} = \frac{XA}{BC'} = \frac{XA}{AH} = \frac{AM}{MC'}$$so $M=P$ which implies $P\in\odot(ABC)$.

This problem was proposed by Burii.

Define $H_b$, $H_c$ as the points symmetric to the sides $AC$ and $AB$ with respect to $H$ respectively. Simple angle chasing shows that $H_bY$ and $H_cX$ are tangent to $(AXY)$. That is enough to imply the result since, $\angle{TPA} = \angle{SPA}$ and $\frown{AT}=\frown{AS}$ in $(ABC)$. or through pascal

Since, $XY$ is not tangent to $\odot (AH)$ $\implies$ $CY \cap \odot (AH)=K$. Let $\Delta DEF$ be the orthic triangle $\implies$ $\angle XAK$ $=$ $\angle BXH-90^{\circ}$ $=$ $\frac{A}{2}$ $\implies$ $AK$ bisects $\angle BAC$ and $\Delta XAK$ $\cong $ $\Delta YAK$. Let $L$ be midpoint of arc $BC$ not containing $A$. Let $E',F'$ be reflection of $H$ over $E,F$ $\implies$ $\angle XF'F=\angle XHF$ $=$ $\angle BAL$ $=$ $\angle LF'F$ $\implies$ $X$ $\in$ $LF'$ and similarly, $Y$ $\in$ $LE'$. Also, $L$ $\in$ $AK$. Hence, If $AL$ $\cap$ $BE$ $=$ $M$ $\implies$ $BXML$ is cyclic $\implies$ $\angle XLA$ $=$ $90^{\circ}$ $-$ $A$ $\implies$ $\angle LXY$ $=$ $\angle LYX$ $=$ $A$

lemma: let $AXY$ a triangle with $AX=AY$ and B is located on the line $XY$ then the locus of the reflection of$ B$ wrt $AX$ is the tangent through$ X$ to $(AXY)$ proof: while the locus of $B$ is a line the the locus of $B'$ is also a line. let$ B=X $the$ B'=X$ and let$ B=Y$ the its clear that $B'$ is on the tangent then the line is the tangent through $X$ now apply our lemma onto $\triangle AXY$ wrt $H$ the $H_B$ and $H_C $are on $X$-tangent and $Y$-tangent it suffices to show that$ AH_BH_CP $are cyclic which is trivial since$\angle H_BPH_C=\angle XPY=180-A$

$P$ is anti-Steiner point of $XY$ wrt triangle $ABC$ => $Q.E.D$

Bascially the above post but with some dtails: $\angle PYX=\angle A$ and $\angle XYA=90^{\circ}-\angle A/2$ so $\angle PYC=90^{\circ}-\angle A/2=\angle XYA$ which means $PY$ is the reflection of $XY$ in $CA$.Similarly $PX$ is the reflection $XY$ in $AB$.So $P$ is anti-Steiner point of $XY$ wrt triangle $ABC$.So $P$ lies in the circumcircle of $ABC$

similar config : https://artofproblemsolving.com/community/c6h89098p519896

Reflect the Orthocenter across $AC, AB$ and name the points $H_B, H_C$ analogously. These points lie on the circumcircle of $\triangle ABC$. $\angle H_BYA = \angle HYX = \angle XYA = \angle AXY$, since $\triangle AXY$ is isosceles. This implies that $H_B, Y, P$ are collinear, thus $H_C, X, P$ are collinear similarly. Now It suffices to prove that $H_B, A, H_C, P$ are concyclic. This is trivial enough, as $\angle AH_BP = \angle AH_BY = \angle AHY = 180 - \angle AHX = 180 - \angle AXH_C = 180 - \angle AH_CP$ Q.E.D

ryan17 wrote: Let $ABC$ be an acute triangle. Points $X$ and $Y$ lie on the segments $AB$ and $AC$, respectively, such that $AX=AY$ and the segment $XY$ passes through the orthocenter of the triangle $ABC$. Lines tangent to the circumcircle of the triangle $AXY$ at points $X$ and $Y$ intersect at point $P$. Prove that points $A, B, C, P$ are concyclic. Easy to see that $\angle (\overline{AB}, \overline{XY}) = \angle (\overline{XY}, \overline{XP}) = \frac{\angle BAC}{2}$ and similarly $\angle (\overline{AC}, \overline{XY}) = \angle (\overline{XY}, \overline{YP}) = \frac{\angle BAC}{2}$. Therefore $P$ is the Anti-Steiner Point of $\overline{XY}$ wrt $\triangle ABC$ which means $P \in \odot(ABC)$ as desired.

Let H1 and H2 be reflections of H across the AB and AC. We know AH1BCH2 is cyclic so we can prove AH1PH2 is cyclic in order to solve the problem. ∠H1AH2 = 2∠A and ∠XPY = 180 - 2∠A so we're Done.

Let $H_B$ and $H_C$ be the reflections of $H$ in $\overline{AC}$ and $\overline{BC},$ respectively. Notice $H_B$ lies on $\overline{PY}$ since $$\angle AXH_C=\angle AXH=\angle AXY=\angle AYX.$$Also, $$\angle APH_C=\angle H_BPA=90-\angle XYP=90-\angle BAC=\angle ACH_C.$$$\square$

Let $H$ be the orthocenter of $ABC$, $E = BH \cap AC$, $F = CH \cap AB$, the midpoint of minor arc $BC$ be $M_a$, the $A$-antipode wrt $(ABC)$ be $A_1$, and $N = AP \cap XY$. Because $AX = AY$, we know $A, N, P, M_a$ are collinear along the internal bisector of $\angle BAC$. Since $BCEF$ is cyclic with diameter $BC$, we have $AENHF \sim ABM_aA_1C$, so $$\frac{AN}{AM_a} = \frac{AE}{AB} = \cos{A}.$$Now, observe that the isosceles condition gives $$\angle AXN = \frac{180^{\circ} - \angle XAY}{2} = \frac{\angle B + \angle C}{2}.$$Thus, the Ratio Lemma yields $$\frac{NA}{AP} = \frac{XN}{XP} \cdot \frac{\sin{NXA}}{\sin{AXP}} = \cos{NXP} \cdot \frac{\sin \left(\frac{\angle B + \angle C}{2} \right)}{\sin \left(\frac{\angle B + \angle C}{2} + \angle NXP \right)}$$$$= \cos{A} \cdot \frac{\sin \left( \frac{\angle B + \angle C}{2} \right)}{\sin \left( \frac{\angle B + \angle C}{2} + \angle A \right)} = \cos{A} = \frac{NA}{AM_a}$$so $AP = AM_a$. It follows that $P \equiv M_a$, which finishes. $\blacksquare$ Better: Just notice that $\overline{AXB}$ is the external bisector of $\angle NXP$, so $\frac{NA}{AP} = \frac{XN}{XP}$.