Therefore, $\angle IGP = \angle IEP = 90^\circ$, that is, $IG \perp PG$. Hence, points $P$, $S$ and $T$ are collinear.

Line $PST$ intersects $\triangle ABC$. By Menelaus' Theorem, we have
$$\frac{AS}{SC} \cdot \frac{CP}{PB} \cdot \frac{BT}{TA} = 1.$$
Since $AS = AT$, $CS = CD$ and $BT = BD$, we have
$$\frac{PC}{PB} \cdot \frac{BD}{CD} = 1. \qquad \textcircled{1}$$
Let the extension of $BN$ intersect $PE$ at point $H$. Then line $BFH$ intersects $\triangle PDE$. By Menelaus' Theorem,
$$\frac{PH}{HE} \cdot \frac{EF}{FD} \cdot \frac{DB}{BP} = 1.$$
Since $CF$ is parallel to $BE$, $\frac{EF}{FD} = \frac{PC}{CD}$, we have
$$\frac{PH}{HE} \cdot \frac{PC}{CD} \cdot \frac{DB}{BP} = 1. \qquad \textcircled{2}$$
By ① and ②, we have $PH = HE$. Hence, $PH^2 = HE^2 = HM \cdot HN$. Thus, we have $\frac{PH}{HM} = \frac{HN}{PH}$, $\triangle PHN \sim \triangle MHP$ and $\angle HPN = \angle HMP = \angle NEQ$. Further, since $\angle PEN = \angle EQN$, therefore $\angle ENP = \angle ENQ$. $\square$