Let $E$ be the intersection of line segments $AO$ and $DP$. Also, let $H$ be the foot of the perpendicular from $A$ to $BC$. I claim that triangles $AEP$ and $AHC$ are congruent. It is well known (and easy to check) that $AO$ and $AH$ are isogonal lines. Therefore, $AD$ is the bisector line of angle $\angle EAH$. As a result, $H$ is the reflection of $E$ over line $AD$. This implies $$ AE=AH \quad (1) $$ Again, since lines $AE$ and $AH$ are isogonal, then $$ \angle PAE=\angle HAC \quad (2) $$ And, from the heading, $$ \angle AEP=\angle CHA \quad (3) $$From $(1), (2)$ and $(3)$ we get that triangles $\triangle AEP$ and $\triangle AHC$ are congruent, and the problem has been solved.

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