Concatenation of empty language with any language will give the empty language and ${L_1}^ * = \phi^* = \epsilon$.
Therefore,
$L_1L_2^* \cup L_1^* $
$=\phi.(L_2)^* \cup \phi^ *$
$= \phi \cup \{\epsilon\} \left(\because \phi \text{ concatenated with anything is } \phi \text{ and }\phi^* = \{\epsilon\} \right) $
$= \{\epsilon \} $.
Hence, option (A) is true.
PS: $\phi^* = \epsilon$, where $\epsilon$ is the regular expression and the language it generates is $\{\epsilon\}$.