Answer to the question is (A) $L_1 = L_2$.
Reason is for any PDA which accepts by final state there is an equivalent PDA (equivalent means that accepts the same language) which accepts by empty stack and vice-verse.
Now, this is not the case for DPDAs.
The set of languages accepted by a DPDA by empty stack is a strict subset of the set of languages accepted by a DPDA by final state.
It can also be said that set of languages accepted by a DPDA by empty stack is the set of languages accepted by a DPDA by final state and which has the prefix property.
A language has prefix property means if $w \in L$, then no proper prefix of $w \in L$.
From the above definition of prefix property it must be clear why DPDA by empty stack has this property. If any prefix of a word $w$ ($w$ in $L$) is in $L$ means the stack should have been empty even before completely processing $w$. But, being a deterministic PDA, once the stack becomes empty, the DPDA accepts and halts. So, in no way can a DPDA accepts $w$ and its prefix.
PS: A DPDA with acceptance by empty stack cannot even accept all regular languages- example $a^*$.
Good read: http://www.cs.ucr.edu/~jiang/cs150/slides4week7_PDA+EquivToCFG.pdf