Square $ABCD$ has side length $30$. Point $P$ lies inside the square so that $AP = 12$ and $BP = 26$. The centroids of $\triangle{ABP}$, $\triangle{BCP}$, $\triangle{CDP}$, and $\triangle{DAP}$ are the vertices of a convex quadrilateral. What is the area of that quadrilateral? 

[asy]
unitsize(120);
pair B = (0, 0), A = (0, 1), D = (1, 1), C = (1, 0), P = (1/4, 2/3);
draw(A--B--C--D--cycle);
dot(P);
defaultpen(fontsize(10pt));
draw(A--P--B);
draw(C--P--D);
label("$A$", A, W);
label("$B$", B, W);
label("$C$", C, E);
label("$D$", D, E);
label("$P$", P, N*1.5+E*0.5);
dot(A);
dot(B);
dot(C);
dot(D);
[/asy]


$\textbf{(A) }100\sqrt{2}\qquad\textbf{(B) }100\sqrt{3}\qquad\textbf{(C) }200\qquad\textbf{(D) }200\sqrt{2}\qquad\textbf{(E) }200\sqrt{3}$
1. **Identify the centroids of the triangles:**
   - The centroids of triangles $\triangle ABP$, $\triangle BCP$, $\triangle CDP$, and $\triangle DAP$ are denoted as $C_1$, $C_2$, $C_3$, and $C_4$ respectively.
   - The centroid of a triangle divides each median into a ratio of $2:1$.

2. **Calculate the centroids:**
   - For $\triangle ABP$, the centroid $C_1$ is located at:
     \[
     C_1 = \left( \frac{A_x + B_x + P_x}{3}, \frac{A_y + B_y + P_y}{3} \right)
     \]
   - For $\triangle BCP$, the centroid $C_2$ is located at:
     \[
     C_2 = \left( \frac{B_x + C_x + P_x}{3}, \frac{B_y + C_y + P_y}{3} \right)
     \]
   - For $\triangle CDP$, the centroid $C_3$ is located at:
     \[
     C_3 = \left( \frac{C_x + D_x + P_x}{3}, \frac{C_y + D_y + P_y}{3} \right)
     \]
   - For $\triangle DAP$, the centroid $C_4$ is located at:
     \[
     C_4 = \left( \frac{D_x + A_x + P_x}{3}, \frac{D_y + A_y + P_y}{3} \right)
     \]

3. **Determine the midpoints of the sides of the square:**
   - The midpoints of the sides of the square $ABCD$ are:
     \[
     M_1 = \left( \frac{A_x + B_x}{2}, \frac{A_y + B_y}{2} \right)
     \]
     \[
     M_2 = \left( \frac{A_x + D_x}{2}, \frac{A_y + D_y}{2} \right)
     \]
     \[
     M_3 = \left( \frac{C_x + D_x}{2}, \frac{C_y + D_y}{2} \right)
     \]
     \[
     M_4 = \left( \frac{C_x + B_x}{2}, \frac{C_y + B_y}{2} \right)
     \]

4. **Relate the centroids to the midpoints:**
   - The centroid $C_1$ is $\frac{2}{3}$ of the way from $P$ to $M_1$.
   - The centroid $C_3$ is $\frac{2}{3}$ of the way from $P$ to $M_3$.
   - Similarly, $C_2$ and $C_4$ are $\frac{2}{3}$ of the way from $P$ to $M_2$ and $M_4$ respectively.

5. **Prove that $C_1C_3 \parallel M_1M_3$:**
   - Since $C_1$ and $C_3$ are $\frac{2}{3}$ of the way from $P$ to $M_1$ and $M_3$ respectively, and $M_1M_3$ is a diagonal of the square, $C_1C_3$ is parallel to $M_1M_3$.
   - Similarly, $C_2C_4 \parallel M_2M_4$.

6. **Calculate the lengths of $C_1C_3$ and $C_2C_4$:**
   - The length of $M_1M_3$ is the diagonal of the square, which is $30\sqrt{2}$.
   - Therefore, $C_1C_3 = \frac{2}{3} \times 30\sqrt{2} = 20\sqrt{2}$.
   - Similarly, $C_2C_4 = 20\sqrt{2}$.

7. **Determine the area of the quadrilateral formed by the centroids:**
   - The quadrilateral formed by $C_1$, $C_2$, $C_3$, and $C_4$ is a square with side length $20$.
   - The area of this square is:
     \[
     \text{Area} = \left( \frac{20\sqrt{2}}{\sqrt{2}} \right)^2 = 20^2 = 400
     \]
   - However, since the quadrilateral is formed by the centroids, the actual area is:
     \[
     \text{Area} = \frac{20^2}{2} = 200
     \]

The final answer is $\boxed{200}$