Let $M$ be the midpoint of $\overline{AI}$. Let $N$ be the foot from $I$ to $\overline{PF}$. Thus, as $\measuredangle IEP=90^\circ=\measuredangle INP=\measuredangle IMP$, we get that $EINMP$ is cyclic. Let $R$ be the intersection of $\overline{QE}$ and the incircle of $\triangle ABC$. As $\overline{QF}$ is tangent to $(FNI)$, we get that $QN\cdot QI=QF^2=QR\cdot QE$, we get that $R$ lies on $(EINP)$. Therefore, $$\measuredangle ERM=\measuredangle EIA=\measuredangle ERF,$$which implies that $F,R,M$ are collinear. As $P$ is the intersection of the perpendicular bisector of segment $\overline{AI}$ and $\overline{AC}$, we get that $PI\parallel AB$ as $\measuredangle QAI=\measuredangle IAP=\measuredangle PIA$. Hence, $$\measuredangle AEN=\measuredangle PEN=\measuredangle PIN=\measuredangle PIQ=\measuredangle FQN=\measuredangle AQN,$$which implies that $AQNE$ is cyclic. Claim: $\triangle QNR\sim\triangle ANP$. Proof. Indeed, $\measuredangle NQR=\measuredangle NQE=\measuredangle NAE=\measuredangle NAP$ and $\measuredangle APN=\measuredangle EPN=\measuredangle ERN=\measuredangle QRN$. The claim follows. Now, \begin{align*} \measuredangle ENA=\measuredangle ENP+\measuredangle PNA=\measuredangle RNQ+90^\circ+\measuredangle EPI=\measuredangle RMI+90^\circ+\measuredangle EPI=90^\circ. \end{align*}We are done.

$PA=PI\implies\angle PIA=\angle PAI=\angle IAF$. So $PI\parallel AB$. $QI\perp FP\implies PQ^2-PI^2=FQ^2-FI^2\implies PQ^2-(PE^2+EI^2)=(QI^2-FI^2)-FI^2=(QI^2-EI^2)-EI^2\implies QP^2-EP^2=QI^2-EI^2\implies EQ\perp PI$. But $PI\parallel AB$. Hence $EQ\perp AB$ as desired.