### Instruction: 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:
$\textbf{(A)}\ 1:3 \qquad \textbf{(B)}\ 3:4 \qquad \textbf{(C)}\ 4:3 \qquad \textbf{(D)}\ 3:1 \qquad \textbf{(E)}\ 7:1$.### Response: This problem involves some angle bisector theorems and ratio properties.
Let's start by drawing a diagram of the situation.
[Diagram of $\triangle ABC$ with $AC:CB=3:4$ and exterior angle bisector at $C$ intersecting $BA$ extended at $P$]
The key fact we need to use is that the angle bisector of an exterior angle of a triangle divides the opposite side externally in the same ratio as the adjacent sides.
In other words, since $CP$ is the bisector of the exterior angle at $C$, we have $PA:AB=AC:CB=3:4$.
This means that the correct answer is (B) 3:4.