Square $ ABCD$ has sides of length $ 4$, and $ M$ is the midpoint of $ \overline{CD}$. A circle with radius $ 2$ and center $ M$ intersects a circle with raidus $ 4$ and center $ A$ at points $ P$ and $ D$. What is the distance from $ P$ to $ \overline{AD}$?
[asy]unitsize(8mm);
defaultpen(linewidth(.8pt));
dotfactor=4;

draw(Circle((2,0),2));
draw(Circle((0,4),4));
clip(scale(4)*unitsquare);
draw(scale(4)*unitsquare);
filldraw(Circle((2,0),0.07));
filldraw(Circle((3.2,1.6),0.07));
label("$A$",(0,4),NW);
label("$B$",(4,4),NE);
label("$C$",(4,0),SE);
label("$D$",(0,0),SW);
label("$M$",(2,0),S);
label("$P$",(3.2,1.6),N);[/asy]$ \textbf{(A)}\ 3 \qquad \textbf{(B)}\ \frac {16}{5} \qquad \textbf{(C)}\ \frac {13}{4} \qquad \textbf{(D)}\ 2\sqrt {3} \qquad \textbf{(E)}\ \frac {7}{2}$
1. **Identify the coordinates of key points:**
   - Let the square \(ABCD\) be positioned in the coordinate plane with \(A = (0, 4)\), \(B = (4, 4)\), \(C = (4, 0)\), and \(D = (0, 0)\).
   - The midpoint \(M\) of \(\overline{CD}\) is \((2, 0)\).

2. **Define the circles:**
   - Circle centered at \(M\) with radius 2: \((x - 2)^2 + y^2 = 4\).
   - Circle centered at \(A\) with radius 4: \(x^2 + (y - 4)^2 = 16\).

3. **Find the intersection points of the circles:**
   - Since \(D\) is one of the intersection points, we need to find the other intersection point \(P\).

4. **Set up the equations for the circles:**
   - Circle at \(M\): \((x - 2)^2 + y^2 = 4\).
   - Circle at \(A\): \(x^2 + (y - 4)^2 = 16\).

5. **Solve the system of equations:**
   - Expand and simplify the equations:
     \[
     (x - 2)^2 + y^2 = 4 \implies x^2 - 4x + 4 + y^2 = 4 \implies x^2 - 4x + y^2 = 0 \quad \text{(1)}
     \]
     \[
     x^2 + (y - 4)^2 = 16 \implies x^2 + y^2 - 8y + 16 = 16 \implies x^2 + y^2 - 8y = 0 \quad \text{(2)}
     \]

6. **Subtract equation (1) from equation (2):**
   \[
   (x^2 + y^2 - 8y) - (x^2 - 4x + y^2) = 0 \implies -8y + 4x = 0 \implies 4x = 8y \implies x = 2y
   \]

7. **Substitute \(x = 2y\) into equation (1):**
   \[
   (2y)^2 - 4(2y) + y^2 = 0 \implies 4y^2 - 8y + y^2 = 0 \implies 5y^2 - 8y = 0 \implies y(5y - 8) = 0
   \]
   - Thus, \(y = 0\) or \(y = \frac{8}{5}\).

8. **Find corresponding \(x\) values:**
   - For \(y = 0\), \(x = 0\) (which is point \(D\)).
   - For \(y = \frac{8}{5}\), \(x = 2 \cdot \frac{8}{5} = \frac{16}{5}\).

9. **Coordinates of \(P\):**
   - \(P = \left(\frac{16}{5}, \frac{8}{5}\right)\).

10. **Calculate the distance from \(P\) to \(\overline{AD}\):**
    - The distance from \(P\) to \(\overline{AD}\) is the y-coordinate of \(P\), which is \(\frac{8}{5}\).

The final answer is \(\boxed{\frac{8}{5}}\).