Ooption B is correct.
$L = \{a^lb^mc^n \mid l ≠ m \text{ or } m ≠ n\}$
$(q_0,a,Z_0) \rightarrow (q_0,aZ_0)$
($q_0,a,a) \rightarrow (q_0,aa)$
$(q_0,b,a) \rightarrow (q_1,\epsilon), (q_2,ba) $
[here it is NPDA where we have to check $l\neq m$ or $m\neq n$; for $l\neq m$ we need to pop $a$ for $b$; for $m\neq n$ we need to keep $b$ in stack so that we can pop $b$ for $c$ ]
$(q_1,b,a) \rightarrow (q_1,\epsilon)$
$(q_1,c,a) \rightarrow (q_f,\epsilon)$
$(q_1,b,Z_0) \rightarrow (q_f,\epsilon)$
$(q_2,b,b) \rightarrow q_2,bb)$
$(q_2,c,b) \rightarrow (q_3,\epsilon)$
$(q_3,c,b) \rightarrow (q_3,\epsilon)$
$(q_3,c,a) \rightarrow (q_f,\epsilon)$
$(q3,\epsilon,b) \rightarrow (q_f,\epsilon)$
(A) is wrong as it is not context free
(D) $a^*b^*$ is regular, so must have DFA , and so an equivalent DPDA
(C) can be accepted using DPDA