My solution : Let $ O $ be the circumcenter of $ \triangle ABC $ . Since $ \angle PBC+\angle PCB=\angle EAD+\angle FAD=\angle BAC $ , so $ \angle BPC=180^{\circ}-\angle BAC $ $ \Longrightarrow B, C, H, P $ are concyclic , hence $ HP \parallel BC \Longleftrightarrow \angle HCB=\angle BCP \Longleftrightarrow \angle BAH=\angle CAD \Longleftrightarrow O \in AD $ . Q.E.D

ComplexPhi wrote: Let $ABC$ be an acute triangle , with $AB \neq AC$ and denote its orthocenter by $H$ . The point $D$ is located on the side $BC$ and the circumcircles of the triangles $ABD$ and $ACD$ intersects for the second time the lines $AC$ , respectively $AB$ in the points $E$ respectively $F$. If we denote by $P$ the intersection point of $BE$ and $CF$ then show that $HP \parallel BC$ if and only if $AD$ passes through the circumcenter of the triangle $ABC$. easy problem for a TST!