Observe $\angle AEF=\angle APF=\angle FBC$ so $BCEF$ is cyclic. Thus, $\angle PBA=\angle PCA$. Likewise $\angle PAC=\angle PBC$ and $\angle PAB=\angle PCB$. Let $Q$ be the isogonal conjugate of $P$ in $\triangle ABC$. Then $QA=QB=QC$ hence $P=H$ is the orthocenter of $\triangle ABC$. Now $AE-HE=AF-HF$ by Pithot's Theorem in $AEPF$. Also $AE^2+HE^2=AF^2+HF^2=AH^2$ by Pythagoras Theorem. Solving yields that $\overline{AH}$ bisects angle $EAF$. Now apply this cyclically: $H$ coincides with the incenter so $ABC$ is equilateral and $P$ is its center.

My solution: Note that $\angle BFC = \angle ADC = 180^{\circ}-\angle BEC$ and $\angle AEF = \angle APF = \angle ABD$ $\Rightarrow BCEF$ is cyclic $\Rightarrow \angle BFC = \angle BEC = 90^{\circ} \Rightarrow P$ is the orthocenter of $\triangle ABC$. Now, Let $AP \cap EF = K$ and $T$ be the center of $\odot (AEPF) \Rightarrow K, T, I$ is collinear, where $I$ is the incenter of $AEPF$. $\Rightarrow I$ lies on $AP \Rightarrow AP$ bisects $\angle EAF$ and $\angle EPF$. Thus, $\angle AEF = \angle APF = \angle APE = 90^{\circ}-\angle EAP \Rightarrow AK \perp EF \Rightarrow EF \parallel BC$ But, $EF$ is antiparallel to $BC \Rightarrow \triangle ABC$ is isosceles with $AB = AC$. As the situation is symmetric, we get that $\triangle ABC$ is equilateral with $P$ as its orthocenter (i.e. its center).

Just note that by PoP, we have $BCEF$ cyclic, and that $\angle AFP = \angle PDB = \angle PEC$, so $\angle BEC = 90^\circ$. So $P$ is the orthocenter. Also, by Pitot on $ABHC$, we have that the orthocenter is an equal detour point of $ABC$, which happens only if $ABC$ is equilateral. The rest is clear.

How is this a TST problem??