[TIKZ]\draw[thin] (0,0) circle (5cm);
\coordinate[label=left:B] (B) at (-4,3);
\coordinate[label=right:C] (C) at (6,3);
\coordinate[label=above:A] (A) at (-0.33,7);
\coordinate[label=above: P] (P) at (-1,4.9);
\coordinate[label=left: G] (G) at (-5,-0);
\coordinate[label=right: F] (F) at (3.8,-3.25);
\coordinate[label=above: D] (D) at (-0.27,3);
\coordinate[label=below: E] (E) at (2.2,3);
\draw (B) -- (C)-- (A)-- (B);
\draw (B) -- (P)-- (C)-- (B);
\draw (B) -- (G)-- (F)-- (B);
\draw (P) -- (F);
\draw (A) -- (E);
\draw (A) -- (P);
\draw (G) -- (P);
\draw (C) -- (F);
\begin{scope}
\path[clip] (B) -- (A) -- (P) -- cycle;
\draw[thick,red,double] (A) circle (1.3295);
\end{scope}
\begin{scope}
\path[clip] (B) -- (G) -- (P) -- cycle;
\draw[thick,red,double] (G) circle (1.3295);
\end{scope}
\begin{scope}
\path[clip] (E) -- (A) -- (C) -- cycle;
\draw[thick,red,double] (A) circle (1.3295);
\end{scope}
\begin{scope}
\path[clip] (B) -- (F) -- (P) -- cycle;
\draw[thick,red,double] (F) circle (1.3295);
\end{scope}
\begin{scope}
\path[clip] (A) -- (B) -- (P) -- cycle;
\draw[thick,blue] (B) circle (0.4177);
\end{scope}
\begin{scope}
\path[clip] (B) -- (P) -- (G) -- cycle;
\draw[thick,blue] (P) circle (0.4177);
\end{scope}
\begin{scope}
\path[clip] (A) -- (C) -- (P) -- cycle;
\draw[thick,blue] (C) circle (0.4177);
\end{scope}
\begin{scope}
\path[clip] (B) -- (F) -- (G) -- cycle;
\draw[thick,blue] (F) circle (0.4177);
\end{scope}
[/TIKZ]
Let $ABFC$ and $ABGP$ be parallelograms. Since $\triangle AEC \cong \triangle FDB$, we have $\angle BFP=\angle BFD=\angle CAE=\angle BAP=\angle BGP$. This implies $BGFP$ are concyclic.
Since $\triangle APC \cong \triangle BGF$, we have $\angle ABP=\angle BPG=\angle BFG=\angle ACP$ (Q.E.D.).
Remark: I know the parallelogram $ABFC$ looks distorted, but please cut me some slack (LOL), I have been trying my best to draw the diagram in TiKZ and some of my initial attempts work some never (which puzzled me much). I will continue to investigate the whys and eventually I will replace this diagram with a perfect one. Thanks to Klaas van Aarsen for his patience and constant help to guide me through some of the issues I encountered.