Let $\triangle ABC$ be a scalene triangle. Point $P$ lies on $\overline{BC}$ so that $\overline{AP}$ bisects $\angle BAC.$ The line through $B$ perpendicular to $\overline{AP}$ intersects the line through $A$ parallel to $\overline{BC}$ at point $D.$ Suppose $BP=3$ and $PC=5.$ What is $AD?$

Let $BD$ meet $AP$ at $E$ and $AC$ at $F$. Triangles $ABE$ and $AFE$ share $AE$, have right angles at $E$ and equal angles at $A$, so $\triangle ABE\cong\triangle AFE$ and $AF=AB$. By the angle bisector theorem, $AB:AC=BP:PC=3:5$, so $AF:AC=3:5$, hence $AF:FC=3:2$.
Since $AD\parallel BC$, $\triangle ADF\sim \triangle CBF$, so $AD:BC=AF:FC=3:2$, and $BC=BP+PC=8$, so
$$AD=\frac32(8)=\boxed{12}.$$