In triangle $ABC$ lines $CE$ and $AD$ are drawn so that 

$\dfrac{CD}{DB}=\dfrac{3}{1}$ and $\dfrac{AE}{EB}=\dfrac{3}{2}$. Let $r=\dfrac{CP}{PE}$ 

where $P$ is the intersection point of $CE$ and $AD$. Then $r$ equals:

[asy]
size(8cm);
pair A = (0, 0), B = (9, 0), C = (3, 6);
pair D = (7.5, 1.5), E = (6.5, 0);
pair P = intersectionpoints(A--D, C--E)[0];
draw(A--B--C--cycle);
draw(A--D);
draw(C--E);
label("$A$", A, SW);
label("$B$", B, SE);
label("$C$", C, N);
label("$D$", D, NE);
label("$E$", E, S);
label("$P$", P, S);
//Credit to MSTang for the asymptote
[/asy]



$\textbf{(A)}\ 3 \qquad
\textbf{(B)}\ \dfrac{3}{2}\qquad
\textbf{(C)}\ 4 \qquad
\textbf{(D)}\ 5 \qquad
\textbf{(E)}\ \dfrac{5}{2}$