$ADEF$ is cyclic, so $\measuredangle AEF=\measuredangle ADF=CBA$. Thus $\triangle AEF \stackrel{-}{\sim} \triangle ABC$, and the conclusion follows since the orthocenter and circumcenter are isogonal conjugates.

As $\angle DFA = \angle DEA = 90^{\circ}$ so $FD \parallel PE$ and $DE \parallel FP$. Finally \[\angle PAE = \angle PFE = \angle FED = \angle FAD = \angle BAD = 90^{\circ} - \angle ABC = \angle OAC \blacksquare \]