Let $k$ be the circumcircle of the quadrangle $ABCD$ and $k_1$, $k_2$ the circumcircles of the triangles $PAB$ and $PCD$, respectively. In the interior of the angle $BPC$, consider the half-line $PT$ such that $|\angle BPT| = |\angle BAP|$. Then the hypothesis on $P$ implies that (Fig. 1)
$$|\angle TPC| = |\angle BPC| - |\angle BPT| = |\angle BPC| - |\angle BAP| = |\angle PDC|.$$
Thus $PT$ is the common interior tangent of the circles $k_1$ and $k_2$.

Fig. 1

Let us first assume that the sides $AB$ and $CD$ of the given cyclic quadrangle are not parallel. Since the segments $AB$ and $CD$ are common chords of the circles $k_1$, $k$ and $k_2$, $k$, respectively, there exists a unique point $Q$ having the same power with respect to all three circles $k$, $k_1$ and $k_2$. The point $Q$ is the intersection of the three lines $AB$, $DC$ and $PT$. Without loss of generality, we can assume that the point $Q$ is located on the half-line $BA$ beyond the point $A$ (Fig. 1). Then we have
$$|\angle QPA| = |\angle PBA| \quad (1)$$
Since $|\angle AEP| = |\angle AFP| = 90^\circ$, the quadrangle $AEPF$ is cyclic, and
$$|\angle FEP| = |\angle FAP| = |\angle DAP| \quad (2)$$
Similarly we see that the quadrangle $DGPF$ is cyclic. It follows that
$$|\angle BPC| = |\angle BAP| + |\angle PDC| = |\angle EFP| + |\angle PFG| = |\angle EFG| \quad (3)$$
Since further $|\angle PEQ| = |\angle PGQ| = 90^\circ$, the quadrangle $QEPG$ is also cyclic and
$$|\angle GEP| = |\angle GQP| = |\angle DQP| \quad (4)$$
From the relations (2), (4) and the equality $|\angle DAP| + |\angle QPA| = |\angle QDA| + |\angle DQP|$, we further have
$$|\angle FEG| = |\angle FEP| - |\angle GEP| = |\angle DAP| - |\angle DQP| = |\angle QDA| - |\angle QPA| \quad (5)$$
Since the quadrangle $ABCD$ is cyclic, $|\angle QDA| = |\angle QBC|$. From the relations (1) and (5) we thus get
$$|\angle FEG| = |\angle QBC| - |\angle PBA| = |\angle PBC|$$
Using finally the relations (3) and (6) we see that the triangles $FEG$ and $PBC$ are similar (as they have two congruent angles).

An analogous argument can be used when the point $Q$ is located on the half-line $AB$ beyond the point $B$. If the lines $AB$ and $CD$ are parallel, then $ABCD$ is an equilateral trapezoid, with bases $AB$ and $CD$. Since the points $E$, $P$, $G$ are collinear and the common interior tangent of the circles $k_1$ and $k_2$ is parallel to both lines $AB$ and $CD$, the triangles $APD$ and $PBC$ are congruent. The similarity of the triangles $EFG$ and $APD$ thus implies also the similarity of the triangles $EFG$ and $PBC$.

This completes the proof.