Points $ K$, $ L$, $ M$, and $ N$ lie in the plane of the square $ ABCD$ so that $ AKB$, $ BLC$, $ CMD$, and $ DNA$ are equilateral triangles. If $ ABCD$ has an area of $ 16$, find the area of $ KLMN$.

[asy]unitsize(2cm);
defaultpen(fontsize(8)+linewidth(0.8));
pair A=(-0.5,0.5), B=(0.5,0.5), C=(0.5,-0.5), D=(-0.5,-0.5);
pair K=(0,1.366), L=(1.366,0), M=(0,-1.366), N=(-1.366,0);
draw(A--N--K--A--B--K--L--B--C--L--M--C--D--M--N--D--A);
label("$A$",A,SE);
label("$B$",B,SW);
label("$C$",C,NW);
label("$D$",D,NE);
label("$K$",K,NNW);
label("$L$",L,E);
label("$M$",M,S);
label("$N$",N,W);[/asy]

$ \textbf{(A)}\ 32 \qquad \textbf{(B)}\ 16 \plus{} 16\sqrt {3} \qquad \textbf{(C)}\ 48 \qquad \textbf{(D)}\ 32 \plus{} 16\sqrt {3} \qquad \textbf{(E)}\ 64$
1. Given that the area of square \(ABCD\) is 16, we can find the side length of the square. Since the area of a square is given by \(s^2\), where \(s\) is the side length, we have:
   \[
   s^2 = 16 \implies s = \sqrt{16} = 4
   \]

2. Consider the equilateral triangle \(AKB\). Since \(AKB\) is equilateral, all its sides are equal to the side length of the square, which is 4. We draw the altitude from \(K\) to \(AB\), which splits \(AKB\) into two 30-60-90 right triangles.

3. In a 30-60-90 triangle, the ratio of the sides opposite the 30°, 60°, and 90° angles is \(1 : \sqrt{3} : 2\). Therefore, the altitude from \(K\) to \(AB\) (which is the side opposite the 60° angle) is:
   \[
   \text{altitude} = \frac{4 \sqrt{3}}{2} = 2\sqrt{3}
   \]

4. The total length of the diagonal of \(KLMN\) is the sum of the side length of the square and twice the altitude of the equilateral triangles (one altitude on each side of the square). Thus, the diagonal length is:
   \[
   4 + 2 \times 2\sqrt{3} = 4 + 4\sqrt{3}
   \]

5. The area of a square can also be calculated using the length of its diagonal \(d\) with the formula:
   \[
   \text{Area} = \frac{d^2}{2}
   \]
   Substituting \(d = 4 + 4\sqrt{3}\), we get:
   \[
   \text{Area} = \frac{(4 + 4\sqrt{3})^2}{2}
   \]

6. Expanding \((4 + 4\sqrt{3})^2\):
   \[
   (4 + 4\sqrt{3})^2 = 16 + 32\sqrt{3} + 48 = 64 + 32\sqrt{3}
   \]

7. Therefore, the area of \(KLMN\) is:
   \[
   \text{Area} = \frac{64 + 32\sqrt{3}}{2} = 32 + 16\sqrt{3}
   \]

The final answer is \(\boxed{32 + 16\sqrt{3}}\)