By angle chasing $\angle{AXP} = \angle{ACB}$ and $\angle{AYC} = \angle{ABC} \implies X, Y \in (BC) \implies P$ is the orthocenter of the triangle formed by $BC, BY, CX$, which finishes.

Concurrency of Altitudes in any triangle will solve the problem…

Let $BY\cap AC=T$ Just to prove $P$ is the orthocenter of the $\triangle TBC$ Note that $\angle AYP=\angle AFP=\angle ABC\Rightarrow BY\bot YC$ Similar,$BX\bot XC$ So $P$ is the orthocenter of the $\triangle TBC$

How can PD intersect both AB and AC

I claim CX and BY are altitudes in triangle PBC. Since AEPX and ACED are cyclic, by radical center so is CDPX. Therefore <CXP=<CDP=90deg. BY being an altitude is similar, so the concurrency point is just the orthocenter of PBC

We first establish the following key claim. Claim : Points $X$ and $Y$ lie on the circle $(ABC)$. Proof : Observe that $(AFPY)$ and $(AFDB)$ are clearly cyclic (one is given the other is due to right angles). Note that \[CP \cdot CY = CF \cdot CA = CD\cdot CB\]Thus, $(BDPY)$ must be cyclic. Thus, \[\measuredangle CYB = \measuredangle PYB = \measuredangle PDB = 90^\circ\]Thus, $Y$ indeed lies on $(ABC)$. Similarly, $X$ lies on $(ABC)$ as well, as claimed. Now, simply note that by Radical Center on circles $(ABC)$, $(BDPY)$ and $(CDPX)$ we conclude that lines $PD$, $BY$ and $CX$ concur as desired.

First we claim $X$ and $Y$ lie on $(ABC)$. Let $X'$ and $Y'$ be the intersections of $BP$ and $CP$ with $ABC$. We get $$\angle AEP=\angle AED=90^{\circ} -\angle ABC=\angle ACB=\angle AXB=\angle AXP$$$$\angle AFP=\angle AFD=90^{\circ} -\angle ACB=\angle ABC=\angle AYC=\angle AYP$$This implies that $APXE$ and $APYF$ are concyclic so $X=X'$ and $Y=Y'$ proving the claim. Now let $BY$ and $CX$ intersect at $G$. We get that $P$ is the orthocenter of $GBC$ so $G$, $P$, and $D$ are collinear. The result follows.

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Soaeb wrote: How can PD intersect both AB and AC

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