We only show the proof in the case when $E \in (AC)$, $F \in (AB)$, the other cases being similar (the diagram on the left shows such a case).
From $\angle PBC + \angle PCB = \angle EAD + \angle FAD = \angle BAC$ follows $\angle BPC = 180^\circ - \angle BAC = \angle BHC$, hence $B, C, H, P$ are concyclic.

Therefore $HP \parallel BC$ is equivalent to $B, C, P, H$ being the vertices of an isosceles trapezoid, which translates into $\angle HCB = \angle PBC$. But $\angle HCB = \angle HAB$ and $\angle PBC = \angle DAE$, hence $\angle HCB = \angle PBC \Leftrightarrow \angle BAH = \angle CAD$, which means that $AH$ and $AD$ are isogonals, i.e. $AD$ passes through the circumcenter of $ABC$.

Another solution:

As above, $\angle BPC = 180^\circ - \angle BAC = \angle BHC$. This means that the reflections, $H'$ and $P'$, of $H$ and $P$, respectively, across $BC$ are on the circumcircle of $ABC$. Moreover, $\angle BCF = \angle BAD = \angle BED$, hence $CDPE$ is cyclic. Then $\angle CDP = \angle AEB = \angle ADB$, which means that $P' \in AD$.

Now $HP \parallel BC \Leftrightarrow H'P' \parallel BC$, which is equivalent to $H'P' \perp AH$, i.e. $\angle AH'P' = 90^\circ$, and the conclusion follows readily.