Let $H=AP\cap BC$, since $PFBH$ and $AEPF$ is cyclic we get $\measuredangle ABC=\measuredangle FPA=\measuredangle FEA$ $\Longrightarrow$ $BCEF$ is cyclic, let $A'$ the antipode of $A$ wrt $\omega$ since $PF\parallel A'B$ and $PE\parallel CA'$ we get the perpendicular bisectors of $BF$ and $CE$ intersect in $PA'$ be that point $O$ $\Longrightarrow$ $O$ is the circumcenter of $\odot (BCEF)$ $\Longrightarrow$ $O,P, A'$ are collinear, it so easy note that $G=\overline {OPA'}\cap \omega$. On the other hand by radical axis in $\odot (AFE)$, $\odot (BCEF)$ and $\omega$ we get $AG$, $FE$, $BC$ are concurrent in $X$, let $Y$ $=$ $BE$ $\cap$ $CF$ $\Longrightarrow$ by Brocard's theorem in $\odot (BCEF)$ we get $OY$ $\perp$ $AX$, but $OG$ $\perp$ $AY$ hence $O$ $,$ $G$ $,$ $Y$ are collinear $\Longrightarrow$ $O$ $,$ $P$ $,$ $G$ $,$ $A'$ $,$ $Y$ are collinear hence $GP$, $BE$ and $CF$ are concurrent in $Y$.

This is a very common miquel configuration. $\angle AEP + \angle AFP = 90^{\circ}+90^{\circ} = 180^{\circ}$ impies $AEPF$ cyclic. Similarly, we get $PDCE$ and $PDBF$ cyclic where $D$ is the foot of the perpendicular from $A$ to $BC$. From this we get $BCEF$ cyclic and $G$ its miquel point. Let $BE \cap CF = X$. Also, let the center of $BCEF$ be $O$. From the miquel configuration, we have $G, X, O$ colinear. So, we reduce the problem into showing $X, P, O$ colinear. $\angle FEP = \angle FEB - \angle PEX = \angle FCB - 90^{\circ} + \angle BEC = \angle FCB - \angle BCO = \angle FCO$. So, $EP$ meets $CO$ on the circle. Let this point be $Y$. Analogously, $FP$ meets $BO$ at $Z$ which is also on the circle. Now we apply pascal's theorem to hexagon $BZFCYE$. We get our conclusion.

Cool problem. Similar solution as above. Firstly we have that $$\angle PEA =\angle PFA = 90^\circ$$from where we obtain that $PEFA$ is cyclic. Let $H$ be feet of altitude from $A$. Now from $$\angle PFB = \angle PHB = 90^\circ$$and $$\angle PEC = \angle PHC = 90^\circ$$we obtain that $PFBH$ and $PEHC$ are cyclic too. Now we are doing some angle chase. $$\angle FBC = 180^\circ - \angle FPH =\angle FPA = \angle FEA = 180^\circ - \angle FEC$$With this we get that $BCFE$ is cyclic. Now let $T = BE \cap CF$. Let $O$ be circumcenter of $BCEF$. $G$ is Miquel point of the quadrilateral and the rest is the same as above so there is no point in writing it all over again

See my solution on my Youtube channel here. It is very similar to FabrizioFelen's solution, but with a slight difference in the middle (I think his way is actually simpler). https://www.youtube.com/watch?v=Rl6S_sU7rK4

You can work this up using Hong Kong 2001 Test 1 Q1. $\definecolor{A}{RGB}{0,90,255}\color{A}\fbox{Solution}$ As I proved there, if you denote by $O$ the circumcenter of quadrilateral $BFEC$ then points $G,P,O$ lie on line perpendicular to $AG$. Now take $X$ as the intersection of lines $BE,CF$. By radical axis theorem and Brocard's theorem $OX$ is the line perpendicular to $AG$. There's only one line perpendicular to $AG$ passing through point $O$, thus $X\in GP.\blacksquare$ #1802

$BEFC$ is cyclic.So from radical center $AG,EF,BC$ concur at $S$. If $BE,CF$ concur at $T$ then $GT,GP$ are perpedicular to $SA$ so $G,T,P$ collinear so we are done