Since $\angle AFO=\angle AEB=180^\circ-\angle DPA$ so $F$ is the circumcenter of $\triangle ADP$. Other hand, $\angle EPD=\angle ECD=\angle EAB$ so $A,P,D,Q$ are concyclic implies $F$ is the circumcenter of $\triangle APQ$ i.e $FP=FQ$

$APDQ$ cyclic + $PE \perp AD \implies$ $F$ is the center of $APD$

Assume wlog that $AB<AC$. We first observe that $\angle QAD=\angle BAD=\angle BCD=\angle ECD=\angle EPD=\angle QPD$ and so $A,P,D,Q$ are concyclic. Secondly, $\angle EPC=180^{\circ}-\angle EDC=180^{\circ}-\angle ABC$ yields that $B,P,C,Q$ lie on a circle. Let $F^*$ denote the circumcenter of the triangle $APQ$. Clearly, $F^*$ lies on the $A$ altitude in $ABC$ and $F^*P=F^*Q$. Thus, we only need to show that $OF^* \parallel PQ$. Notice that $\angle AOF^*=90^{\circ}$ since $OF^*$ is the perpendicular bisector of $AD$. Now, $\angle F^*AO=\angle (OF^*,BC)=\angle B-\angle C=\angle PEC$ which yields that $\angle (OF^*,BC)=\angle (PQ,BC)$ and so $OF^* \parallel PQ$. The result follows.

Note that by angle chase $APDQ$ and $BPCQ$ are cyclic. Now $FO \perp AD$ and bisects it $\Longrightarrow$ circumcenter of $APDQ \in OF$ and since $AF, AE$ are isogonal in $\triangle APQ$ circumcenter lies on $AF$ and thus the circumcenter actualy is $F$ and thus $FP = FQ$ holds. Does T5 mean a hard problem? I don't think that this is hard or even a medium problem.

Ig this solution is bit different. $\angle EPC=180°-\angle EDC=180°-\angle ABC=\angle QBC$ Hence, $QBPC$ is a cyclic quadrilateral. By PoP, $PE.EQ=BE.EC=AE.ED$ $\implies PE.EQ=AE.ED \implies \text{APDQ is cyclic quadrilateral}$ Also simlpe angle chasing gives $F$ is circumcenter of $\Delta APD$ Hence $F$ is the circumcenter of $(APDQ)$. Result immediate.