* Ta có: $x^{2}-2 x-3=0 \Leftrightarrow (x+1)(x-3)=0 \Leftrightarrow \left[\begin{array}{c}x=-1 \\ x=3\end{array}\right.$ $\Rightarrow E=\{-1;3\}$
* Lại có: $(x+1)(2 x-3)=0 \Leftrightarrow \left[\begin{array}{c}x=-1 \\ x=\frac{3}{2}\end{array}\right.$ $\Rightarrow G=\{-1;\frac{3}{2}\}$
Vậy $P=E \cap G=\{-1\}$.