Kronecker product of $M$ and $N$. Then $$PP^*P=p^{n+m+s_1+s_2}P$$ holds.

Let $P=M \otimes N$. Then $$\begin{split}
    PP^*P&=\left( M \otimes N\right) \left( M^* \otimes N^*\right)\left( M \otimes N\right)   \\
    &=\left( MM^*M\right) \otimes \left( NN^*N \right)\\
    &=\left( p^{n+s_1}M\right) \otimes \left( p^{m+s_2}N \right)\\
    &=p^{n+m+s_1+s_2} \left( M\otimes N\right) \\
    &=p^{n+m+s_1+s_2}P\\
    \end{split}$$

Let $f: \mathbb{F}_{p^{n}} \to \mathbb{F}_{p} $ and $g : \mathbb{F}_{p^{m}} \to \mathbb{F}_{p}$ be two $s_1$-plateaued and $s_2$-plateaued functions respectively. Then there exists an $(s_1+s_2)$-plateaued function from $\mathbb{F}_{p^{n+m}}$ to $\mathbb{F}_{p}$.

Let $M$ and $N$ be matrices associated with $f$ and $g$ such that $m_{x,y}=\zeta_p^{f(x+y)}$ and $n_{a,b}=\zeta_p^{g(a+b)}$. Let $P=M \otimes N$. We need to show that there exist a function $h$ such that the entries of $P$ can be associated with $h$.

We will index the rows and columns of $P$ by the elements of $\mathbb{F}_{p^{n+m}}$. First note that there is a subgroup $H$ of the additive group of $\mathbb{F}_{p^{n+m}}$ which is isomorphic to the additive group of $\mathbb{F}_{p^{m}}$. Let us fix a transversal $T=\{\gamma_1,\gamma_2,\dots, \gamma_{p^n} \}$ of this subgroup in $\mathbb{F}_{p^{n+m}}$. Now order the rows and columns of $P$ by $\gamma_1+H$, $\gamma_2+H$, ..., $\gamma_{p^n}+H$ in the block form. Here we have isomorphisms namely $$\phi_1: H \to \mathbb{F}_{p^{m}}$$ and $$\phi_2: \{\gamma_1+H,\gamma_2+H,\dots, \gamma_{p^n}+H \} \to \mathbb{F}_{p^{n}}$$

If 