Note that if $M$ is the midpoint of $BC$ then $EM \perp AD$ since $EM$ is isogonal to the $E$-altitude in $\triangle{BEC}$, so $QR \parallel EM$. Consider the homothety at $Y = BC \cap EF$ taking $QR$ to $EM$; it suffices to show that this homothety takes $P$ to the point $S = EF \cap (BC)$, because since $M$ is the center of $(BC)$, that would imply $PR = QR$. Thus, it suffices to show that $\frac{YP}{YS} = \frac{PQ}{ES}$. Since $PQ = \frac{PE^2}{PF}$, this simplifies to $\frac{YP}{YS} = \frac{EP}{ES} \cdot \frac{PE}{PF}$. $\angle{ESC} = \angle{EBC} = \angle{EAF}$, so $FASC$ is cyclic $\implies ES \cdot EF = EA \cdot EC \implies \frac{EP}{ES} = \frac{EF}{EH}$. Also, by Power of a Point at $Y$ onto $(BC), (ABCD)$, $\frac{YP}{YS} = \frac{YE}{YH}$, if $H = \overrightarrow{EF} \cap (ABCD)$, so it suffices to show that $\frac{YE}{YH} = \frac{EF}{EH} \cdot \frac{PE}{PF}$, or $EY \cdot EH \cdot PF = EF \cdot EP \cdot YH$, which simplifies to $EP \cdot EH \cdot |EF - EY| = EF \cdot EY \cdot |EH - EP|$. By Power of a Point at $Y$, $$YB \cdot YC = (EP - EY)(EH + EY) = EP \cdot EH + EY(EP - EH) - EY^2$$$$\implies EP - EH = \frac{YB \cdot YC + EY^2 - EP \cdot EH}{EY} = \frac{EY^2\left(\frac{YB \cdot YC}{EY^2}+1\right) - EP \cdot EH}{EY}.$$So, $$EF \cdot EY \cdot (EP - EH) = EF\left(EY^2\left(\frac{YB \cdot YC}{EY^2}+1\right) - EP \cdot EH\right)$$and it suffices to show that that expression is equal to $EP \cdot EH \cdot (EY - EF)$. Cancelling the $-EF \cdot EP \cdot EH$ from both sides, this becomes $$EP \cdot EH = EF \cdot EY\left(\frac{YB \cdot YC}{EY^2}+1\right),$$and since $EP \cdot EH = ES \cdot EF$ by Power of a Point, it suffices to show that $ES = EY\left(\frac{YB \cdot YC}{EY^2}+1\right)$. However, by Law of Sines in $\triangle{YEB}, \triangle{YEC}$, $\frac{YB \cdot YC}{EY^2} = \frac{BS \cdot SC}{EB \cdot EC} = \frac{[SBC]}{[EBC]} = \frac{SY}{EY}$, so it suffices to show that $\frac{ES}{EY} = \frac{SY}{EY} + 1$, which is obvious since $SY + EY = ES$. $\square$

I’ll just post the original graph

Attachments:

Here is a sketch of my solution. Let $O, M, N$ be the circumcenters of $\odot (ABCD)$, $\odot (ADE)$, $\odot (BCE)$ respectively. Also let $K=PE \cap BC$, and $P_1,K_1,F_1$ be the another intersection of line $PE$ and $\odot (ABCD)$, $\odot (ADE)$, $\odot (BCE)$ respectively. 1.It is not difficult to show that $\triangle OMK_1 \cong \triangle F_1NO$, from which we can get $PF_1=P_1K_1$. 2.Apply an inversion at $E$ with power Ψ which indicates the power of $E$ with respect to $\odot (ABCD)$. Denote by $Q_1$ the image of $Q$ under such inversion. We can prove that $PE^2=PQ \cdot PF$ is equivalent to $EQ_1=EP_1+EF_1$, and $KE^2=KQ \cdot KP_1$ is equivalent to $EQ_1=EK_1+EP$ as well, hence $PE^2=PQ \cdot PF$ and $KE^2=KQ \cdot KP_1$ are equivalent. 3.Let $L$ be the another intersection of line $BC$ and $\odot (BEP)$. Redefne $R$ as the circumcenter of $\triangle PQL$. Since $EL \parallel P_1C$, therefore $QL \parallel EC$. Also by some easy angle chasing we may have $R$ lies on $BC$ and $QR \perp AD$. Hence the new definition of $R$ is equivalent to the original one, which implies $RP=RQ$ as desired.