Create NFA for the given expression and from that grammer is as follows:
replace A0 with S and A1 with A
S->0S | 1A
A -> 0A | 1A | epsilon
the above grammer doesn't match with option A and B
In the options A and B they have production rule A0 -> A1 (As mentioned by @vishal8492) which will produce string "0" which is not accepted by the regular expression.
So the answer would D.