Let $N,P_1,P_2$ be submodules of $M$, such that \begin{align} P_2 \subseteq P_1, && N+P_1 = N+ P_2, && N\cap P_1 = N\cap P_2. \end{align} Show that $P_1 = P_2$.
Proof:
Let $p_1\in P_1$. Since $P_1 \subset N+ p_1 = N+ P_2$ there exist $n,p_2$ such that $p_1 = n+ p_2$, so $p_1-p_2 =n$.
Since $P_2 \subseteq P_1$, $n = p_1 - p_2 \in P_1\cap N = P_2 \cap N$, so $n\in P_2$. We conclude $p_1\in P_2$.
Qn: Is it possible to say anything if we do not assume $P_2\subseteq P_1$? For example, the sum $N+P_1$ is direct?
Your proof is fine.
Note that it is not true if we do not assume $P_2\subset P_1$. For then we can consider $M=\mathbb R^2$, $N=\mathbb R\times\{0\}$, $P_1=\{0\}\times\mathbb R$, and $P_2=\{(x,x)\in\mathbb R^2\}$. Then $N+P_1 = N+ P_2$ and $ N\cap P_1 = N\cap P_2$, but $P_1\cap P_2=\{0\}.$