smc

Let $D$ be the origin. $A$ is the point $(0,4)$ and $M$ is the point $(2,0)$. We are given the radius of the quarter circle and semicircle as $4$ and $2$, respectively, so their equations, respectively, are: $x^2+(y-4{)}^2=4^2$ $(x-2{)}^2+y^2=2^2$ Subtract the second equation from the first: $x^2+(y-4{)}^2-(x-2{)}^2-y^2=12$ $4x-8y+12=12$ $x=2y.$ Then substitute: $(2y{)}^2+(y-4{)}^2=16$ $4y^2+y^2-8y+16=16$ $5y^2-8y=0$ $y(5y-8)=0.$ Thus $y=0$ and $y=\frac{8}{5}$ making $x=0$ and $x=\frac{16}{5}$. The first value of $0$ is obviously referring to the x-coordinate of the point where the circles intersect at the origin, $D$, so the second value must be referring to the x coordinate of $P$. Since $\overline{AD}$ is the y-axis, the distance to it from $P$ is the same as the x-value of the coordinate of $P$, so the distance from $P$ to $\overline{AD}$ is $\frac{16}{5}\Rightarrow \boxed{\frac{16}{5}}.$