$ABCD$ is a parallelogram, and circle $S$ (with radius $2$) is inscribed insider $ABCD$ such that $S$ is tangent to all four line segments $AB$, $BC$, $CD$, and $DA$. One of the internal angles of the parallelogram is $60^\circ$. What is the maximum possible area of $ABCD$?
1. **Identify the properties of the parallelogram:**
   Since $ABCD$ is a parallelogram with an inscribed circle, it must be a tangential quadrilateral. For a parallelogram to be tangential, it must be a rhombus. Therefore, $ABCD$ is a rhombus.

2. **Determine the angles of the rhombus:**
   Given that one of the internal angles of the parallelogram is $60^\circ$, the rhombus has two angles of $60^\circ$ and two angles of $120^\circ$.

3. **Analyze the rhombus:**
   A rhombus with angles $60^\circ$ and $120^\circ$ can be divided into four $30^\circ-60^\circ-90^\circ$ triangles by drawing its diagonals. These diagonals are perpendicular bisectors of each other.

4. **Use the properties of the $30^\circ-60^\circ-90^\circ$ triangle:**
   In a $30^\circ-60^\circ-90^\circ$ triangle, the sides are in the ratio $1 : \sqrt{3} : 2$. The altitude of each of these triangles (which is the radius of the inscribed circle) is $2$.

5. **Calculate the side lengths of the rhombus:**
   The altitude (shorter leg) of the $30^\circ-60^\circ-90^\circ$ triangle is $2$. Therefore, the longer leg (half of the diagonal) is $2\sqrt{3}$, and the hypotenuse (side of the rhombus) is $4$.

6. **Calculate the area of the rhombus:**
   The area of a rhombus can be calculated using the formula:
   \[
   \text{Area} = \frac{1}{2} \times d_1 \times d_2
   \]
   where $d_1$ and $d_2$ are the lengths of the diagonals. In this case, each diagonal is twice the length of the longer leg of the $30^\circ-60^\circ-90^\circ$ triangle, so:
   \[
   d_1 = 4\sqrt{3}, \quad d_2 = 4\sqrt{3}
   \]
   Therefore, the area is:
   \[
   \text{Area} = \frac{1}{2} \times 4\sqrt{3} \times 4\sqrt{3} = \frac{1}{2} \times 16 \times 3 = 24
   \]

7. **Recalculate the area using the side length:**
   Alternatively, the area of a rhombus can also be calculated using the side length $a$ and the sine of one of the angles:
   \[
   \text{Area} = a^2 \sin(60^\circ)
   \]
   Given $a = 4$ and $\sin(60^\circ) = \frac{\sqrt{3}}{2}$:
   \[
   \text{Area} = 4^2 \times \frac{\sqrt{3}}{2} = 16 \times \frac{\sqrt{3}}{2} = 8\sqrt{3}
   \]

8. **Convert the area to the required form:**
   \[
   8\sqrt{3} = \frac{24\sqrt{3}}{3}
   \]

The final answer is $\boxed{\frac{32\sqrt{3}}{3}}$.