1963 AMC 12 Question 39

In triangle $ABC$ lines $CE$ and $AD$ are drawn so that $\dfrac{CD}{DB}=\dfrac31$ and $\dfrac{AE}{EB}=\dfrac32.$

Let $r=\dfrac{CP}{PE},$ where $P$ is the intersection point of $CE$ and $AD.$ Then $r$ equals: