Two circles with radius $2$ and radius $4$ have a common center at P.  Points $A, B,$ and $C$ on the larger circle are the vertices of an equilateral triangle.  Point $D$ is the intersection of the smaller circle and the line segment $PB$. Find the square of the area of triangle $ADC$.
1. **Identify the given information and the goal:**
   - Two concentric circles with radii 2 and 4 centered at point \( P \).
   - Points \( A, B, \) and \( C \) form an equilateral triangle on the larger circle.
   - Point \( D \) is the intersection of the smaller circle and the line segment \( PB \).
   - We need to find the square of the area of triangle \( \triangle ADC \).

2. **Determine the side length of the equilateral triangle \( \triangle ABC \):**
   - Since \( A, B, \) and \( C \) lie on the larger circle with radius 4, the side length \( a \) of the equilateral triangle can be found using the formula for the side length of an equilateral triangle inscribed in a circle:
     \[
     a = 2R \sin 60^\circ = 2 \cdot 4 \cdot \frac{\sqrt{3}}{2} = 4\sqrt{3}
     \]

3. **Calculate the area of \( \triangle ABC \):**
   - The area \( A \) of an equilateral triangle with side length \( a \) is given by:
     \[
     A = \frac{\sqrt{3}}{4} a^2 = \frac{\sqrt{3}}{4} (4\sqrt{3})^2 = \frac{\sqrt{3}}{4} \cdot 48 = 12\sqrt{3}
     \]

4. **Determine the position of point \( D \):**
   - Point \( D \) is the intersection of the smaller circle (radius 2) and the line segment \( PB \). Since \( PB \) is a radius of the larger circle, \( PB = 4 \).
   - \( D \) is halfway between \( P \) and \( B \), so \( PD = 2 \).

5. **Calculate the height \( ED \) of \( \triangle ADC \):**
   - Drop a perpendicular from \( P \) to \( AC \) and let the intersection be \( E \). Since \( \triangle ABC \) is equilateral, \( E \) is the midpoint of \( AC \).
   - The height \( PE \) of \( \triangle ABC \) can be calculated as:
     \[
     PE = \sqrt{4^2 - \left(\frac{4\sqrt{3}}{2}\right)^2} = \sqrt{16 - 12} = 2
     \]
   - Since \( D \) is halfway between \( P \) and \( B \), \( ED = \frac{2}{3} \cdot EB \).

6. **Calculate the area of \( \triangle ADC \):**
   - The area of \( \triangle ADC \) is \( \frac{2}{3} \) of the area of \( \triangle ABC \):
     \[
     \text{Area of } \triangle ADC = \frac{2}{3} \cdot 12\sqrt{3} = 8\sqrt{3}
     \]

7. **Find the square of the area of \( \triangle ADC \):**
   - The square of the area of \( \triangle ADC \) is:
     \[
     (8\sqrt{3})^2 = 64 \cdot 3 = 192
     \]

The final answer is \(\boxed{192}\)