jmc

We place the points on a coordinate system: $D$ at the origin, $C$ and $A$ on the positive $x$- and $y$-axes, respectively. Then the circle centered at $M$ has equation $$(x-2{)}^2+y^2=4$$and the circle centered at $A$ has equation $$x^2+(y-4{)}^2=16.$$Solving these equations for the coordinates of $P$ gives $x=16/5$ and $y=8/5$, so the answer is $\boxed{16/5}$.



We also could have solved this problem with a little trigonometry:

Let $\angle MAD=\alpha$. Then $$\begin{array}{rl}PQ& =(PA)\sin (\angle PAQ)\\ & =4\sin (2\alpha )\\ & =8\sin \alpha \cos \alpha \\ & =8\left(\frac{2}{\sqrt{20}}\right)\left(\frac{4}{\sqrt{20}}\right)\\ & =\boxed{\frac{16}{5}}.\end{array}$$