Two squares with side $12$ lie exactly on top of each other.
One square is rotated around a corner point through an angle of $30$ degrees relative to the other square.
Determine the area of the common piece of the two squares.

[asy]
unitsize (2 cm);

pair A, B, C, D, Bp, Cp, Dp, P;

A = (0,0);
B = (-1,0);
C = (-1,1);
D = (0,1);
Bp = rotate(-30)*(B);
Cp = rotate(-30)*(C);
Dp = rotate(-30)*(D);
P = extension(C, D, Bp, Cp);

fill(A--Bp--P--D--cycle, gray(0.8));
draw(A--B--C--D--cycle);
draw(A--Bp--Cp--Dp--cycle);

label("$30^\circ$", (-0.5,0.1), fontsize(10));
[/asy]
1. **Identify the Geometry**:
   - We have two squares with side length \(12\) units.
   - One square is rotated by \(30^\circ\) around a corner point relative to the other square.

2. **Determine the Common Area**:
   - The common area can be divided into two \(30^\circ-60^\circ-90^\circ\) triangles.

3. **Properties of \(30^\circ-60^\circ-90^\circ\) Triangles**:
   - In a \(30^\circ-60^\circ-90^\circ\) triangle, the sides are in the ratio \(1 : \sqrt{3} : 2\).
   - The side opposite the \(60^\circ\) angle (which is the hypotenuse in this case) is \(12\) units.

4. **Calculate the Lengths of the Other Sides**:
   - The side opposite the \(30^\circ\) angle is \(\frac{12}{\sqrt{3}} = 4\sqrt{3}\) units.
   - The side opposite the \(90^\circ\) angle (the hypotenuse) is \(12\) units.

5. **Calculate the Area of One Triangle**:
   - The area of a triangle is given by \(\frac{1}{2} \times \text{base} \times \text{height}\).
   - Here, the base is \(12\) units and the height is \(4\sqrt{3}\) units.
   \[
   \text{Area of one triangle} = \frac{1}{2} \times 12 \times 4\sqrt{3} = 24\sqrt{3} \text{ square units}
   \]

6. **Calculate the Total Area of the Common Region**:
   - Since there are two such triangles, the total area is:
   \[
   \text{Total area} = 2 \times 24\sqrt{3} = 48\sqrt{3} \text{ square units}
   \]

The final answer is \(\boxed{48\sqrt{3}}\)