Answer
$a=\frac{L}{3}$
Work Step by Step
$
\begin{array}{cc}
↺+\Sigma M_A=0 ; & \\-P\left(\frac{2 L}{3}-a\right)+C_y(L-a)+P a=0 \\
C_y=\frac{2 P\left(\frac{L}{3}-a\right)}{L-a} \\
↺+\Sigma M=0 ; \quad M=\frac{2 P\left(\frac{L}{3}-a\right)}{L-a}\left(\frac{L}{3}\right)=0 \\
2 P L\left(\frac{L}{3}-a\right)=0 \\
a=\frac{L}{3}
\end{array}
$
