Just a really tiring angle chase . Anyway, here's the solution (which easily follows from backtracking the problem): We have $\angle BQP=\angle BAP=\angle QCP$, which means that $BQ$ is tangent to $\odot (QPC)$, i.e. $BQ^2=BP \cdot BC$. Also, $$\angle QDC=\angle CAP=\angle QBP=\angle QBC \Rightarrow BCQD \text{ is cyclic.}$$Now, $$\angle CED=\angle CDE=180^{\circ}-\angle CDB=180^{\circ}-\angle CQB=\angle AQB=\angle APB$$which gives that $CPDE$ is cyclic. Then $BD \cdot BE=BC \cdot BP=BQ^2$. But this translates to $$\angle QED=\angle BQD=\angle BCD=\angle PCD=\angle PED \Rightarrow P,Q,E \text{ are collinear.}$$Thus, we have $$\angle QFD=\angle QEC=\angle PEC=\angle PDC \Rightarrow QF \parallel PD$$Finally, we get that $$\angle FGB=\angle DPB=\angle DEC=\angle EDF \Rightarrow BDFG \text{ is cyclic. } \blacksquare$$

Attachments:

Nice Claim. $BDQC$ is cyclic. $$\measuredangle CDQ=\measuredangle PAC=\measuredangle PAQ=\measuredangle PBQ=\measuredangle CBQ\qquad \square$$Claim. $ABCE$ is cyclic. $$\measuredangle BEC=\measuredangle DEC=\measuredangle CDE=\measuredangle CDB=\measuredangle CQB =\measuredangle AQB=\measuredangle APB=\measuredangle BAC \qquad \square$$Claim. $CPDE$ is cyclic. $$\measuredangle DEC=\measuredangle BEC=\measuredangle BAC=\measuredangle APB=\measuredangle APC =\measuredangle DPC\qquad \square$$Claim. $E,Q,P$ are collinear. $$\measuredangle CPQ=\measuredangle BPQ=\measuredangle BAQ=\measuredangle BAC=\measuredangle DEC =\measuredangle CDE=\measuredangle CPE \qquad \square$$ To finish the problem, we prove that $AP\parallel QF$; $$\measuredangle PDC=\measuredangle PEC=\measuredangle QEC=\measuredangle QFC=\measuredangle GFC\implies AP\parallel QF.$$Now, $$\measuredangle BDF=\measuredangle EDF=\measuredangle EDC=\measuredangle CED=\measuredangle CEB=\measuredangle CAB=\measuredangle BPA=\measuredangle BGF,$$hence we are done.