Saudi Arabian Mathematical Competitions

First, we note that the circle of diameter $TS$ is the Apollonius circle of triangle $ABC$ then
$$\frac{BP}{CP} = \frac{BA}{CA} = \frac{BT}{CT}$$
or
$$\frac{BP}{BA} = \frac{CP}{CA},$$
which implies that the bisector of angle $B$ in triangle $ABP$ and the bisector of angle $C$ in triangle $ACP$ pass through the same point on $AP$. Denote that point as $K$. So $CE$, $BF$, $AP$ are concurrent at $K$.

It is easy to see that $I \in AT$. Consider triangle $APT$ and we have
$$\frac{IA}{IT} \cdot \frac{KP}{KA} \cdot \frac{DT}{DP} = \frac{BA}{BT} \cdot \frac{BP}{BA} \cdot \frac{BT}{BP} = 1$$
then by applying Ceva's theorem, we can see that three lines $AD$, $IP$, $KT$ are concurrent.

Continue, consider triangle $KBC$ and we have
$$\frac{FK}{FB} \cdot \frac{EC}{EK} = \frac{PK}{PB} \cdot \frac{PC}{PK} = \frac{PC}{PB} = \frac{TC}{TB}$$
or
$$\frac{FK}{FB} \cdot \frac{EC}{EK} \cdot \frac{TB}{TC} = 1 \tag{2}$$
then $KT$, $BE$, $CF$ are also concurrent.

Finally, suppose that $BE$, $CF$, $AD$ are concurrent at $X$ then denote $BE \cap CF = X$, we will have $X \in KT$ based on (2), but $X \in AD$ then $X \in KT \cap AD$.
Therefore, based on (1), we have $X \in IP$, then the result will follow.