5. As shown in Figure 1, the convex quadrilateral $ABCD$ is inscribed in a circle, with $CB=CD$. Two moving points $E$ and $F$ are on $AC$ and $AD$ respectively, and satisfy $EF \parallel BD$. Let $BE$ intersect $CF$ at point $P$, then the geometric position of point $P$ is ( ).
(A) outside the circle
(B) inside the circle
(C) on the circle
(D) cannot be determined
5.C.

As shown in Figure 6, let $AC$ intersect $BD$ at $K$. Given $CB = CD$, we know $\angle BAC = \angle CAD$.

Thus, $\triangle ABC \backsim \triangle AKD$.
Therefore, $\frac{AB}{AC} = \frac{AK}{AD} = \frac{AE}{AF}$.
Hence, $\triangle ABE \backsim \triangle ACF$, $\angle 1 = \angle 2$.
At this point, $\angle BPC = 180^{\circ} - (\alpha + \beta + \gamma + \angle 2)$
$$
= 180^{\circ} - (\alpha + \beta + \gamma + \angle 1) = \angle BAC.
$$

Therefore, points $A$, $B$, $C$, and $P$ are concyclic, and point $P$ lies on the circumcircle of $\triangle ABC$.