jmc

Let the altitude from $P$ onto $AE$ at $Q$ have lengths $PQ=h$ and $AQ=r$. It is clear that, for a given $r$ value, $AP$, $BP$, $CP$, $DP$, and $EP$ are all minimized when $h=0$. So $P$ is on $AE$, and therefore, $P=Q$. Thus, $AP$=r, $BP=|r-1|$, $CP=|r-2|$, $DP=|r-4|$, and $EP=|r-13|.$ Squaring each of these gives: $A{P}^2+B{P}^2+C{P}^2+D{P}^2+E{P}^2=r^2+(r-1{)}^2+(r-2{)}^2+(r-4{)}^2+(r-13{)}^2=5r^2-40r+190$ This reaches its minimum at $r=\frac{40}{2\cdot 5}=4$, at which point the sum of the squares of the distances is $\boxed{110}$.