jmc

Since $\angle WAX=90^{\circ }$ regardless of the position of square $ABCD$, then $A$ always lies on the semi-circle with diameter $WX$.

The center of this semi-circle is the midpoint, $M$, of $WX$.

To get from $P$ to $M$, we must go up 4 units and to the left 3 units (since $WX=2$), so $P{M}^2=3^2+4^2=25$ or $PM=5$.

Since the semi-circle with diameter $WX$ has diameter 2, it has radius 1, so $AM=1$.

So we have $AM=1$ and $MP=5$.



Therefore, the maximum possible length of $AP$ is $5+1=\boxed{6}$, when $A$, $M$, and $P$ lie on a straight line.