jmc

First, it is probably a good idea to sketch our triangle: Since $P$ is the intersection of medians, it is the centroid of $△ABC.$ That means that the ratio of $CP:PN$ is $2:1.$ We can easily find $CN$ by using the right triangle $△CBN.$ Since $N$ is the midpoint of $AB,$ $BN=3.$ At this point, we might recognize that $△CBN$ is a $30^{\circ }-60^{\circ }-90^{\circ }$ triangle, which is handy since we can easily find $CN=6.$ Therefore, $CP$ is $\frac{2}{3}$ of $CN,$ or $\boxed{4}.$