smc

Let $AC=3n$ and $BC=4n$. Draw $X$, where $X$ is on $BC$ and $AC\parallel PX$. By AA Similarity, $△ABC\sim △PBX$, so $PX=3an$, $BX=4an$, and $CX=4an-4n$. Also, let $\angle ABC=a$ and $\angle BAC=b$. Since the angles of a triangle add up to $180^{\circ }$, $\angle BCA=180-a-b$. By Exterior Angle Theorem, $\angle ACX=a+b$, and since $CP$ bisects $\angle ACX$, $\angle PCX=\frac{a+b}{2}$. Because $AC\parallel PX$, $\angle BXP=180-a-b$. Thus, $\angle CPX=\frac{a+b}{2}$, making $△CPX$ an isosceles triangle. Because $△CPX$ is isosceles, $PX=CX$, so $4an-4n=3an$. That means $a=4$, so $PB=4\cdot AB$. Thus, $PA=PB-AB=3\cdot AB$, so $PA:AB=3:1$. The answer is $\boxed{3:1}$, and it can be verified (or obtained) by making $ABC$ a 3-4-5 right triangle.