In $\triangle ABC$ the ratio $AC:CB$ is $3:4$. The bisector of the exterior angle at $C$ intersects $BA$ extended at $P$ 
($A$ is between $P$ and $B$). The ratio $PA:AB$ is:
$\textbf{(A)}\ 1:3 \qquad \textbf{(B)}\ 3:4 \qquad \textbf{(C)}\ 4:3 \qquad \textbf{(D)}\ 3:1 \qquad \textbf{(E)}\ 7:1$
Let $AC = 3n$ and $BC = 4n$.  Draw $X$, where $X$ is on $BC$ and $AC \parallel PX$.  By AA Similarity, $\triangle ABC \sim \triangle PBX$, 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 $\triangle CPX$ an [isosceles triangle](https://artofproblemsolving.com/wiki/index.php/Isosceles_triangle).

Because $\triangle CPX$ 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 $\boxed{\textbf{(D)}}$, and it can be verified (or obtained) by making $ABC$ a 3-4-5 right triangle.