In $\triangle ABC$, medians $\overline{AD}$ and $\overline{CE}$ intersect at $P$, $PE=1.5$, $PD=2$, and $DE=2.5$. What is the area of $AEDC?$ 
[asy]
unitsize(75);
pathpen = black; pointpen=black;
pair A = MP("A", D((0,0)), dir(200));
pair B = MP("B", D((2,0)), dir(-20));
pair C = MP("C", D((1/2,1)), dir(100));
pair D = MP("D", D(midpoint(B--C)), dir(30));
pair E = MP("E", D(midpoint(A--B)), dir(-90));
pair P = MP("P", D(IP(A--D, C--E)), dir(150)*2.013);
draw(A--B--C--cycle);
draw(A--D--E--C);
[/asy]

$\textbf{(A)}\ 13 \qquad
\textbf{(B)}\ 13.5 \qquad
\textbf{(C)}\ 14 \qquad
\textbf{(D)}\ 14.5  \qquad
\textbf{(E)}\ 15 $
1. **Identify the given information and the properties of the centroid:**
   - Medians $\overline{AD}$ and $\overline{CE}$ intersect at $P$.
   - $PE = 1.5$, $PD = 2$, and $DE = 2.5$.
   - The centroid $P$ divides each median into a $2:1$ ratio.

2. **Determine the lengths of the medians:**
   - Since $P$ is the centroid, it divides $\overline{CE}$ into $CP:PE = 2:1$. Therefore, $CP = 2 \times 1.5 = 3$.
   - Similarly, $P$ divides $\overline{AD}$ into $AP:PD = 2:1$. Therefore, $AP = 2 \times 2 = 4$.

3. **Calculate the area of quadrilateral $AEDC$:**
   - The area of a quadrilateral with perpendicular diagonals is given by $\frac{1}{2} \times d_1 \times d_2$, where $d_1$ and $d_2$ are the lengths of the diagonals.
   - Here, the diagonals are $\overline{AD}$ and $\overline{CE}$, with lengths $AD = AP + PD = 4 + 2 = 6$ and $CE = CP + PE = 3 + 1.5 = 4.5$.

4. **Apply the formula for the area of the quadrilateral:**
   \[
   \text{Area of } AEDC = \frac{1}{2} \times AD \times CE = \frac{1}{2} \times 6 \times 4.5
   \]
   \[
   = \frac{1}{2} \times 27 = 13.5
   \]

The final answer is $\boxed{13.5}$.