2. The ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)$ intersects the negative half-axis of the $x$-axis at point $C$, and $A$ is a point on the ellipse in the first quadrant. The line $O A$ intersects the ellipse at another point $B$, and the left focus of the ellipse is $P$. Connecting $A P$ intersects $B C$ at point $D$. If $\overrightarrow{C D}=\frac{3}{2} \overrightarrow{D B}$, then the eccentricity of the ellipse is . $\qquad$
2. $\frac{1}{4}$.

From the problem, we know $C(-a, 0), P(-c, 0)$.
Let $A\left(x_{0}, y_{0}\right)\left(y_{0} \neq 0\right)$.
Then $B\left(-x_{0},-y_{0}\right)$.
By $\overrightarrow{C D}=\frac{3}{2} \overrightarrow{D B}$, we have
$$
x_{D}=-\frac{2 a+3 x_{0}}{5}, y_{D}=-\frac{3 y_{0}}{5} \text {. }
$$

Then $k_{A P}=\frac{y_{0}}{x_{0}+c}, k_{P D}=\frac{3 y_{0}}{2 a+3 x_{0}-5 c}$.
By $k_{A P}=k_{P D}$, we get $a=4 c$, i.e., $e=\frac{1}{4}$.