Let L2 = compl(L1). Now, L1 intersect L1 is empty and L1 union L2 is ∑* and hence regular. And L2 here is regular and infinite - so A option is eliminated.
Now, we have L1 union L2 as regular and hence we have a DFA (say D) for it. We also have DFA for L1 (say D1). Now, we can make a DFA for L2 (say D2) by doing D intersect compl(D1), as complement of a regular language is regular and regular set is closed under intersect. So, L2 must be regular.
Now, consider L2 = {}. Now, L1 intersect L2 is empty and L1 union L2 is L1 which is regular. But here, L2 is regular and finite.
Hence option D- none of these.