Q:

How prove this $\max_i|\textbf{p}_i(t)|\leq \max_i|\textbf{p}_i(0)|$

Let 
$$\sum_{j=1}^N a_{ij}=1, \ \ \ \ i=1,\dots,N$$
and consider
$$\frac{d}{dt}\textbf{p}_i=\bar{\textbf{p}}_i-\textbf{p}_i, \ \ \ \bar{\textbf{p}}_i:=\sum_{j=1}^N a_{ij} \textbf{p}_j.$$
Prove that
$$\max_i|\textbf{p}_i(t)|\leq \max_i|\textbf{p}_i(0)|.$$

A:

When it comes to differential inequalities, you need to add Gronwall's inequality to your belt. If
$$ \frac{d}{dt} y \leq  p(t) y$$
then you may conclude (where existence permits)
$$ y(t) \leq y(a) \exp \left (\int_a^t p(s) ds \right)$$
Note that with your condition on $A$ you have
$$\sum_{j=1}^N a_{ij} p_j \leq \max_{i}|p_i|$$ 
Skipping over a few technical details (which you should verify), we'll see (from your original differential inequality)
$$\frac{d}{dt} \max_{i}|p_i| \leq 0$$
Thus
$$\max_{i}|p_i| \leq \max_{i} | p_i(0)| $$

