First we have to add to the statement that $A-X-Y$ are in this order . Let $K$ be the midpoint of $O'X$ we have $N$,the ninepoint circle, is midpoint of $AO'$ thus $ NK \parallel XY=AX $(line ) but $KX=\frac R2,NY=\frac R2$ where $R$ is the radius of $(ABC)$ and $(BHC)$ hence $XNKY$ is isoceles trapezoid therefore the bisector of $XY$ is also the bisector of $NK$ whence $KP=NP=\frac R2$ so $O'XP$ is rectangle triangle . RH HAS

$R_{BHC}=R_{ABC}$. So the radius of nine-point circle is half of the radius of $(BHC)$. Now we use homothety with centre $X$ and scale $2$ and take nine-point circle$(N), P, Y$ to $(N'), P', Y'$ respectively. It's enough to show that $P'$ lies on $(BHC)$, and as $PX=PY$ therefore it's enough to prove that $O'N'||AX$. Let $O$ be the circumcircle of $(ABC)$. $XHN'O$ and $AOO'H$ are parallelograms, so $AO'$ and $XN'$ and $OH$ bisect each other. And $AXO'N'$ is a parallelogram. Thus we're done!

PROF65 wrote: whence $KP=NP=\frac R2$ so $O'XP$ is rectangle triangle . RH HAS Can you please explain how the conclusion follows ?

Didnt solve this back then Let $AP,AY$ hit $\omega$ at $P'$ and $Y'$ respectively such that $AY'>AY$ and $AP'>AP$. Let $XP$ hit $\omega$ at $X'$ again. We wish to show that $X'P=XP$. Consider a homothety at $A$ with ratio $2$. This maps $\Omega \mapsto \omega$. Note that $P\mapsto P'$ and $Y\mapsto Y'$ under this homothety. Hence we have $PY\parallel P'Y'$. And $P'Y'=2PY$. Note that $\angle YXX' = \angle PYX = \angle XY'P'$, so $XX'P'Y'$ is a isosceles trapezoid with $XX'=P'Y'$. Hence we have $XX' = P'Y' = 2PY=2PX$ and we are done. $\square$