Berkeley Math Circle

Solution:

First, from just the fact the three quadrilaterals are cyclic, we contend that $P$ is the orthocenter of $\triangle ABC$. By power of a point, we have $AF \cdot AB = AP \cdot AD = AE \cdot AC$, hence quadrilateral $BFEC$ is cyclic as well.
$$\measuredangle AEP = \measuredangle AFP = \measuredangle BFC = \measuredangle BEC = \measuredangle PEA$$
which implies $\overline{BPE} \perp \overline{AC}$; similarly $\overline{CPF} \perp \overline{AB}$ and so on.

Next we show that if the quadrilaterals also have an incircle, then $P$ is the incenter of $\triangle ABC$. Now consider quadrilateral $PDCE$, which we now know has diameter $\overline{PC}$, hence $PD^2 + CD^2 = PE^2 + CE^2$. Since it has an incircle, we have $PD + CE = PE + CD$. Together, this is enough to imply $PD = PE$ and $CD = CE$ (i.e. that $PDCE$ is a kite). In a similar fashion we get $PD = PE = PF$. So $P$ also is the incenter of $\triangle ABC$.

In conclusion $P$ is both the orthocenter and incenter of $\triangle ABC$ which can only occur if $\triangle ABC$ is equilateral with $P$ its center.