I am a little doubtful about the last lines which I wrote to complete this answer,
Corrected One
\begin{align} \frac{1}{n^2} \text{E} \left[\left(\sum_{i=1}^{n} (h(s_i,p_i) - \text{E}[h])^2\right)\right] \text{E} \left[\left(\sum_{i=1}^{n} z_i^2\right)\right] &\le \frac{1}{n^2} \text{E} \left[\left(\sum_{i=1}^{n} (h(s_i,p_i) - \text{E}[h])^2\right)\right] \left(n\cdot N\right) \\ &\le \frac{1}{n^2} (n \cdot M^2) \cdot (n \cdot N) \\ &\leq M^2 N \end{align}
The first inequality follows from the fact that $\text{E} \left[\left(\sum_{i=1}^{n} z_i^2\right)\right] \le n \cdot N$ (since each $z_i^2$ is bounded above by $N$), and the second inequality follows from the fact that $\text{E} \left[\left(\sum_{i=1}^{n} (h(s_i,p_i) - \text{E}[h])^2\right)\right] \le n \cdot M^2$ (since each $|h(s_i,p_i)|$ is bounded above by $M$).
My question is like the following
if $Y_i = h(s_i,p_i) - \text{E}[h]$, then $Y_i^2 \ge 0$ and $Y_i^2 \le M^2$, so we have:
\begin{align} \text{E} \left[\left(\sum_{i=1}^{n} Y_i^2\right)\right] &= \sum_{i=1}^{n} \text{E}[Y_i^2] \\ &\le \sum_{i=1}^{n} \cdot M^2 \\ &\leq n M^2 \\ \end{align}
So we have $\text{E} \left[\left(\sum_{i=1}^{n} (h(s_i,p_i) - \text{E}[h])^2\right)\right] \le n \cdot M^2$,
Can one confirm this part to make my answer concrete?