2003 AMC 12A Problem 17

Place $D=(0,0),$ $C=(4,0),$ and $A=(0,4).$ The circle centered at $M=(2,0)$ is $(x-2)^2+y^2=4,$ and the circle centered at $A$ is $x^2+(y-4)^2=16.$ Solving these equations gives the intersection $P=\left(\dfrac{16}{5},\dfrac85\right).$ Since $\overline{AD}$ lies on the $y$-axis, the distance from $P$ to $\overline{AD}$ is its $x$-coordinate, $\dfrac{16}{5}.$ Thus, the correct answer is B.