As shown in the figure. Since point $P$ lies on the circumcircle of $\triangle ACD$, and $AC = CD$, we have
$$\angle APF = \angle ADC = \angle CAD = \angle CPD,$$
that is, $PC$ is the external bisector of $\angle APD$. Thus, letting the line $AP$ meet $BC$ at $Q$, $PC$ bisects $\angle QPD$. Also, since point $P$ lies on the circle with $BC$ as diameter, $BP \perp PC$, so $BP$ is the external bisector of $\angle QPD$, from which we obtain
$$\frac{BD}{DC} = -\frac{BQ}{QC}.$$
On the other hand, since $AQ$, $BE$, $CF$ meet at point $P$, by Ceva's theorem we have
$$\frac{BQ}{QC} \cdot \frac{CE}{EA} \cdot \frac{AF}{FB} = 1,$$
hence
$$\frac{BD}{DC} \cdot \frac{CE}{EA} \cdot \frac{AF}{FB} = -\frac{BQ}{QC} \cdot \frac{CE}{EA} \cdot \frac{AF}{FB} = -1$$
and then by Menelaus's theorem, the three points $D$, $E$, $F$ are collinear.