It is proved that PD in D is perpendicular to AB, PE in E is perpendicular to AC, and PF in F is perpendicular to BC.

Then because of the two outer angles MAC of triangle ABC, the bisector of angle NCA intersects with point P,

So PD=PE, PE=PF (the distance from any point on the bisector of the angle to both sides of the angle is equal),

So PD=PF,

So point P is on the bisector of angle B (points with equal distance to both sides of an angle are on the bisector of this angle).