Let $Q'$ be the midpoint of $BC$. Note that $(B, C; D, P) = -1$, so $Q'F^2 = Q'B \cdot Q'C = Q'D \cdot Q'P$. Since $AB > AC$, $D$ is between $Q'$ and $P$, so $Q'$ is on segment $BD$ and $Q'F$ is tangent to $(FDP) \iff \angle Q'FD = \angle EPC$. There can only be one point on segment $BD$ that satisfies the angle condition for $Q$, so $Q' = Q$. $Q$ is the midpoint of $BC$. Applying Brocard's theorem on cyclic quadrilateral $BFEC$ with center $Q$, $Q$ is the orthocenter of triangle $AHP$. Hence, $AQ \perp HP$. Thus, if $AQ \perp OH$, then $P, O, H$ must be collinear.

Can't believe I got this, but a solution without projective geo!: Firstly, $\angle BFD= C$, since $FACD$ is cyclic. Now $\angle QFD= \angle EPC= C-B$, just using angle chasing in $\triangle BFP$. So $BFQ= B$. Now since $\triangle BFC$ is right angled, $Q$, must be its circumcentre using some direct angle chasing. So $Q$ is the midpoint of $BC$. So $AQ \cap OH$ =$G$, which is the centroid, because $AQ$ is the median. Now $\angle QGH =90$, so $GHDQ$ is cyclic. . Similarly, $AFGH$ and $AGHE$ are al;so cyclic. Combining the last 2 results we get that $AFGHE$ is cyclic. Also $F,E,D,Q$ all lie on the nine point circle, so $EFDQ$ is cyclic as well. So by radical axes theorem, $OH, FE, QD$ concur , and since the latter 2 meet at $P$, so clearly $O,H,P$ are collinear. $\blacksquare$