A bit too classical, I think. The proof is structured in several Claims. Claim 1: $D$ is the midpoint of the major arc $BC$. Proof: Easily follows due to spiral similarity reasons and the fact that $BE=CF$ $\blacksquare$ Claim 2: Quadrilateral $EFGJ$ is cyclic. Proof: Note that $PE \cdot PF=PA \cdot PD=PJ \cdot PG,$ as desired $\blacksquare$ Claim 3: $AP$ and $JP$ are the external angle bisectors of angles $\angle EAF$ and $\angle EJF$, respectively. Proof: Note that $AP \equiv AD$ and so the first part of the Claim follows. Moreover, we have that $EFGJ$ is cyclic, and so $\angle EJP=\angle EFG=\angle FEG=\angle FJG,$ hence the second part follows, too $\blacksquare$ Now, to the problem, from Claim 3 we know that $\dfrac{JE}{JF}=\dfrac{PE}{PF}=\dfrac{AE}{AF},$ hence we are done.

Another way: Let X be the intersection of angle A bisector with EF. Prove that AXGP is cyclic. It follows that it’ the apollonius circle.

Another Sketch

Claim 1: $D$ is the midpoint of arc $BC$

Proof

Claim 2:$EJGF$ is cyclic.

Proof

Let the angle bisector of $\measuredangle EAF$ meet $EF$ at $K$ Claim 3: $\frac {JE} {JF}=\frac{AE} {AF}$

Proof

$D$ is the Miquel point of $BCFE$, so it's the center of the spiral similarity sending $\overline{BE}$ to $\overline{CF}$. Since $BE=CF$, this is a rotation, so we also get $DB=DC$ and $DE=DF$, hence $D$ is the midpoint of arcs $BC$ and $EF$ containing $A$. Now let the internal $\angle EAF$-bisector intersect $\overline{EF}$ at $K$ and $(AEF)$ again at $L$, which lies on $\overline{DG}$, and let $\overline{DG}$ intersect $\overline{EF}$ at $M$; note that $\triangle LMK \sim \triangle LAD$. Since $\overline{ADP}$ is an external bisector of $\angle EAF$, by Apollonian circles it suffices to show that $APKJ$ is cyclic. Indeed, we have $$\measuredangle AKP=\measuredangle LKM=\measuredangle ADL=\measuredangle ADG=\measuredangle AJG=\measuredangle AJP.~\blacksquare$$

$GE\cap (ABC)=K,GF\cap (ABC)=L$ If $N$ is the midpoint of arc $BAC$, then $NB=NC,BE=CF,\angle EBN=\angle FCN$ thus, $NEB\sim NFC$ which gives that $N$ is the miquel point of $EFBC\iff D$ is the midpoint of arc $BAC$. Since $PE.PF=PA.PD=PJ.PG\implies EFJG$ are cyclic $\implies J$ is the miquel point of $EFKL$. Since $EK.EG=EB.EA$ and $FL.FG=FC.FA$ and $EB=FC,GE=GF,$ we have $\frac{EK}{FL}=\frac{EA}{FA}$. Also $JEK\sim JFL$ hence $\frac{JE}{JF}=\frac{EK}{FL}=\frac{AE}{AF}$ as desired.$\blacksquare$