Alphametic / Number Puzzle

Since $MPB$ is a 3-digit square, $BE \le 31$. So $B \in \{1, 2, 3\}$. The unit digit of $(BE)^2$ is $B$, so $B$ must be 1 or 0 or 5 or 6, but $B \in \{1,2,3\} \implies B = 1$. $BE$ can be 11 or 19 (since $E \neq B$). If $BE = 11$, $11^2 = 121$ ($M=1=B$, invalid). If $BE = 19$, $19^2 = 361$. Thus $M=3, P=6, B=1$, which are distinct integers. So $M = 3$.