Note that $\angle{PDG}=\angle{PDF}=45^{\circ}$ and $\angle{PEG}=\angle{PEF}=45^{\circ}$. So $P$ is the intersection of angle bisectors of $\angle{GDF}$ and $\angle{GEF}$. Consider the midpoint $P'$ of arc $\overarc{FG}$ respect to circle $(GDEF)$. We get that $P'$ is also the intersection of those bisectors, hence $P\equiv P'$. This gives us $P\in (GDEF)$. Since $\angle{PDG}=\angle{PFG}$, we get $PG=PF$, done.

As $BP=BD$ and $\angle PBD=90^\circ$, we get that $\measuredangle BDP=\measuredangle PDF=45^\circ$. Similarly, $\measuredangle GEP=\measuredangle PEC=45^\circ$. Hence we conclude that $DGEP$ and $DFEP$ are cyclic, therefore $DFEPG$ is cyclic. Now, $\measuredangle GFP=\measuredangle GEP=45^\circ=\measuredangle PDF=\measuredangle PGF$, we are done. [asy][asy] size(9cm);defaultpen(fontsize(10pt));pen org=magenta;pen med=mediummagenta;pen light=pink;pen deep=deepmagenta;pen dark=darkmagenta;pen heavy=heavymagenta; pair O,A,B,C,P,D,E,F,G; O=(0,0);A=dir(110);B=dir(205);C=dir(335);P=-A;path x=circle(B,abs(B-P));path y=circle(C,abs(C-P)); D=intersectionpoints(x,B--3A-2B)[0];E=intersectionpoints(y,A--C)[0]; path r=circumcircle(D,E,P);G=intersectionpoints(r,A--2B-A)[1];F=intersectionpoints(r,A--C)[1];path w=circle(O,1); draw(A--B--C--cycle,deep);draw(w,deep);draw(r,med+dashed);draw(D--B--P,org);draw(P--C--E,light);draw(G--E,deep);draw(D--F,deep);draw(G--B,deep); dot("$A$",A,dir(A));dot("$B$",B,dir(B));dot("$C$",C,dir(C));dot("$P$",P,dir(P));dot("$D$",D,dir(D));dot("$E$",E,dir(E));dot("$F$",F,dir(F));dot("$G$",G,dir(G)); [/asy][/asy]