Sketch: $(EDB)$ is tangent to $(ABC)$ at $B$ so $O,G,B$ are collinear. Similarly, $O,H,C$ are collinear. Then angle chase again to get $O,P,B,C$ are concyclic and $O,P,G,H$ are concyclic too. Angle chase again to get $OHDG$ is parallelogram so $OG=DH=HP$, and we're done. p/s: I don't think that proving $O,P,B,C$ concyclic is necessary... You kinda need to prove it when you're trying to prove $O,P,G,H$ concyclic but like you don't have to tell in the proper solution that $O,P,B,C$ are concyclic. But hey we get an extension that $P$ is the center of spiral similarlity mapping $G\to B$ and $H\to C$.

Attachments:

It is quite similar to a problem which appeared in a Chinese contest in 1997.

Different approach: Let $(BED)\cap AB=Q, (CFD)\cap AC=R$. $QD//AC$ and $RD//AB$, so $(BED)$ and $(CFD)$ are tangent to $(ABC)$ at $B,C$ respectively. Let the intersection of tangent lines from $B, C$ $K$, and $OK\cap BC=S$. By radical axis theorem, $(P,D,K)$ collinear $\therefore KS\cdot KO=KC^2=KD\cdot KP\rightarrow (O,P,S,D)$ concyclic$\rightarrow OP\perp PD$ Since $XY\perp PD$, $OP//XY$

As $E$ is the circumcentre of $\Delta ABD\implies \angle BED=2\angle BAD=\angle BAC=\angle A$. Also, $BEPD$ is cyclic $\implies \angle BED=\angle BPD=\angle A$ Again similarly for $\Delta ACD$, as $F$ is the circumcentre $\implies \angle DFC=\angle A$ $DPFC$ is cyclic $\implies \angle DPC=\angle DFC=\angle A$. Therefore, $\angle BPC=2\angle A$. $O$ is the circumcentre of $\Delta ABC\implies \angle BOC=2\angle A$. Therefore, as $\angle BPC=\angle BOC\implies BPOC$ is a cyclic quadrilateral. By simple angle chasing we get that $\angle OCB=90^{\circ}-\angle A\implies \angle BPO=90^{\circ}+\angle A$. $\implies \angle BPD+\angle DPO=90^{\circ}+\angle A \implies \angle DPO=90^{\circ}$ (as $\angle BPD=\angle A$). $\implies DP\perp OP$ $DP$ is the radical axis of the circumcircles of $\Delta ABD$ and $\Delta ACD$. Therefore, $DP \perp XY$. From the above results, we conclude that $OP\parallel XY$ $\blacksquare$