Mike has two similar pentagons. The first pentagon has a perimeter of $18$ and an area of $8 \frac{7}{16}$ . The second pentagon has a perimeter of $24$. Find the area of the second pentagon.
1. Let \( K_1 \) and \( K_2 \) be the perimeters of the first and second pentagons, respectively. Given:
   \[
   K_1 = 18, \quad K_2 = 24
   \]
   
2. Let \( L_1 \) and \( L_2 \) be the areas of the first and second pentagons, respectively. Given:
   \[
   L_1 = 8 \frac{7}{16} = 8 + \frac{7}{16} = 8 + 0.4375 = 8.4375
   \]

3. Since the pentagons are similar, the ratio of their perimeters is equal to the ratio of their corresponding side lengths. Therefore:
   \[
   \frac{K_1}{K_2} = \frac{18}{24} = \frac{3}{4}
   \]

4. The ratio of the areas of similar figures is the square of the ratio of their corresponding side lengths. Hence:
   \[
   \frac{L_1}{L_2} = \left( \frac{K_1}{K_2} \right)^2 = \left( \frac{3}{4} \right)^2 = \frac{9}{16}
   \]

5. Solving for \( L_2 \):
   \[
   \frac{L_1}{L_2} = \frac{9}{16} \implies L_2 = L_1 \cdot \frac{16}{9}
   \]
   Substituting \( L_1 = 8.4375 \):
   \[
   L_2 = 8.4375 \cdot \frac{16}{9} = 8.4375 \cdot 1.777\overline{7} \approx 15
   \]

The final answer is \( \boxed{15} \).