$A \to bA \mid \varepsilon$
$\therefore \quad A = b^*$
$S \to aSAb \mid \varepsilon$
$\equiv S \to aSb^*b \mid \varepsilon$
$\equiv S \to aSb^+ \mid \varepsilon$
$S = a^n\left(b^+\right)^n, \quad n\geq 0$
$S = a^nb^nb^*, \quad n\geq 0$
$S = a^mb^n, \quad m\leq n$
Hence, option B is correct.