Pumping Lemma for Regular Languages is as follows:
If $A$ is a regular language, then there is a number $p,$ such that if $s$ is any string in $A$ of length at least $p,$ $s$ may be divided into three pieces $x,y,z,$ i.e., $s = xyz,$ such that all of the following hold:
- for each $i \geq 0, xy^i z$ is in $A$
- $|y| > 0$
- $|xy| \leq p$
So, pumping lemma is a necessary condition for a language to be regular. In other words, not satisfying pumping lemma is a sufficient condition for a language to be not regular.
Now, even if a language satisfies pumping lemma it may not be regular.
Reference: https://cs.stackexchange.com/questions/41442/a-non-regular-language-satisfying-the-pumping-lemma
So, correct answer is A and D