It is well-known that the reflection $H'$ of the orthocenter $H$ in the line $AC$ lies on the circumcircle of triangle $ABC$. Hence, the circumcenter of triangle $CAH'$ coincides with the circumcenter of triangle $ABC$. But since $H'$ is the reflection of $H$ in the line $AC$, the triangles $ACH$ and $CAH'$ are symmetric with respect to $BC$, and the circumcenter $O'$ of triangle $ACH$ must be the reflection of the circumcenter of triangle $CAH'$ in the line $BC$, i.e. the reflection of the circumcenter of triangle $ABC$ in the line $CA$.

Now since the quadrilateral $AHPC$ is cyclic and since $H$, $Y$ are the orthocenters of triangles $ABC$, and $APC$, respectively, we have that
$$\angle ABC = 180^\circ - \angle AHC = 180^\circ - \angle APC = \angle AYC.$$

Hence the point $Y$ lies on the circumcircle of triangle $ABC$, and therefore $OC = OY = R$, where $R$ denotes the circumradius of triangle $ABC$.

On the other hand, note that the lines $OX$, $XO'$, $O'O$ are the perpendicular bisectors of the segments $AB$, $AP$, and $AC$, respectively, we get
$$\angle OXO' = \angle BAP = \angle PAC = m(\angle XO'O).$$
Thus $OO' = OX$. Combining this with $OC = OY$ and with the parallelism of the lines $XO'$ and $YC$ (note that these two lines are both perpendicular to $AP$), we conclude that the trapezoid $XYCO'$ is isosceles, and therefore $XY = O'C = OC = R$. This completes our proof. $\square$

Remark. If $ABC$ is right-angled at $A$, then the statement is trivially true if we convene that the circumcenter of $AB$ is the midpoint of $AB$ and that the orthocenter of $AC$ is the midpoint of $AC$. Then, we have that $XY = \frac{1}{2}BC = R$.

Because $ABC$ is acute, $H$ lies inside the triangle. We consider the configuration shown above. (For other possible configurations, it is not difficult to adjust our proof properly.)

Let $O$ and $Z$ denote the circumcenters of triangles $ABC$ and $APC$ respectively. Let $\omega$ and $r$ denote the circumcircle and the circumradius of triangle $ABC$ respectively. We will show that
$$XYCZ \text{ is an isosceles trapezoid with } XY = CZ = r. \qquad (13)$$
Because $X$ and $Z$ are the circumcenters of triangle $APB$ and $APC$, line $XZ$ is the perpendicular bisector of segment $AP$. Because $Y$ is the orthocenter of triangle $APC$, $CY \perp AP$. Hence both lines $XZ$ and $CY$ are perpendicular to line $AP$, implying that $XYZC$ is a trapezoid with $XZ \parallel CY$.

Because $X$ and $O$ are the circumcenters of triangles $APB$ and $ABC$, line $XO$ is the perpendicular bisector of segment $AB$. Because $XO \perp AB$ and $XZ \perp AP$, the acute angles formed by lines $XO$ and $XZ$ is equal to the acute angle formed by lines $AP$ and $AB$; that is, $\angle OXZ = \angle BAP$. Likewise, we can show that $\angle OZX = \angle CAP$. Therefore, we have $\angle OXZ = \angle BAP = \angle CAP = \angle OZX$, implying that $OX = OZ$; that is, $O$ lies on the perpendicular bisector of segment $XZ$.

Because $H$ is the orthocenter of acute triangle $ABC$, $\angle AHC = 180^\circ - \angle ABC$. Because $APHC$ is cyclic, we have $\angle APC = \angle AHC = 180^\circ - \angle ABC$. Now in obtuse triangle $APC$, $\angle AYC = 180^\circ - \angle APC = \angle ABC$. (This relates to the fact of orthocenter group: if one point is the orthocenter of the triangle formed by the other three points, then any of the four point is the orthocenter of the triangle formed by the other three.) In particular, this means that $Y$ lies on $\omega$; that is, $OY = OC = r$. Note that in trapezoid $XYCZ$, the perpendicular bisectors of the bases $YC$ and $XZ$ share a common point $O$. Thus, these two bisectors must coincide; that is, $XYCZ$ is an isosceles trapezoid with $XY = CZ$, establishing the first part of (13).

To complete our proof, it suffices to show that $CZ = r$. Let $Q$ be the reflection of $H$ across line $AC$. It is well known that $Q$ lies $\omega$ (because $\angle ACQ = \angle ACH = 90^\circ - \angle BAC = \angle ABH = \angle ABQ$.) We note that triangle $AQC$ and its circumcenter $O$ and triangle $AHC$ and its circumcenter $Z$ are respective images of each other across line $AC$. In particular, we conclude that $CZ = CO = r$, completing our proof.