Solution:

Let the bisector of $\angle B$ hit $AC$ at $D.$ By the Angle Bisector Theorem, $\frac{AD}{DC} = \frac{AB}{BC} = \frac{80}{45},$ and since $AC = 75,$ we get $AD = 48$ and $CD = 27.$ Triangles $BCD$ and $ACB$ share $\angle C,$ with sides in ratio $\tfrac{45}{75} = \tfrac35,$ so they're similar. That gives $BD = \tfrac35 \cdot 80 = 48,$ which equals $AD.$ So $\triangle ADB$ is isosceles. Writing $\angle DAB = \angle DBA = \theta,$ a short angle chase shows $\angle DPC = \angle DCP,$ so $\triangle CDP$ is isosceles too, with $PD = CD = 27.$ Since $P$ is on segment $BD,$ $BP = BD - PD = 48 - 27 = 21.$ Thus, D is the correct answer.