Find The Angle

In $\triangle ABC,$ $\angle BAC = 65^{\circ}, \angle ABC = 40^{\circ}.$ Let $D,E$ be the feet of perpendiculars on the internal angle bisector of $\angle BAC$ from $B,C$ respectively. Let $P$ be the other point of intersection of the circumcircles of $\triangle ADC$ and $\triangle ABE$ apart from $A.$ Find $\angle PAB$ in degrees.