Let $K$ and $L$ be the points of intersection of the line $AF$ with lines $BC$ and $DE$, respectively (Fig. 6). Then
$$\begin{aligned} \angle AGD &= \angle AGF = \angle AGC + \angle CGF = \angle ABC + \angle CAF \\ &= \angle ABK + \angle KAB = \angle CKA = \angle KLD = 180^\circ - \angle ALD. \end{aligned}$$
Hence the quadrilateral $AGDL$ is cyclic. Interchanging the roles of points $B$ and $C$, points $D$ and $E$ and also points $G$ and $H$, we can similarly prove that the quadrilateral $AHEL$ is cyclic. Thus the circumcircles of triangles $ADG$ and $AEH$ meet at point $L$, i.e., $P = L$. Point $L$ was chosen on the line $AF$.

Fig. 6

Choose a point $Z$ on the same side of the line $AF$ as point $B$ such that $ZF \parallel BC \parallel DE$ (Fig. 7). Then $\angle ZFB = \angle FBC = \angle FAC = \angle FAB = \angle FCB$. Hence $ZF$ is tangent to the circumcircle of the triangle $ABC$ at $F$, implying that $\angle GFZ = \angle GHF = \angle GHE$. As $\angle GFZ = \angle DFZ = \angle FDE = 180^\circ - \angle GDE$, we altogether have $\angle GHE = 180^\circ - \angle GDE$. Consequently, the quadrilateral $DEHG$ is cyclic. Hence $|FD| \cdot |FG| = |FE| \cdot |FH|$, implying that $F$ lies on the radical axis of the circumcircles of triangles $ADG$ and $AEH$. The desired result follows.

Fig. 7