Taking $L$ as a property of r.e. languages, we can have $L_{yes} = \{0, 00, 0000, \dots\}$ and $L_{no} = \{0, 00, 0000, \dots\} \cup \{000\}$. Here $L_{yes} \subset L_{no}$ and hence $L$ is a non-monotonic property. Rice's second theorem is applicable and so $L$ is not r.e. (unrecognizable).
For $\bar L$ also the same thing works. Here the property is "contains the description of $M$ such that if $M$ accepts $w$ and $M$ does not accept $ww$. We can have $L_{yes} = \{0\}$ and $L_{no} = \{0, 00, 000, \dots\} $. Here $L_{yes} \subset L_{no}$ making $\bar L$ a non-monotonic property.
So, both $L$ and $\bar L$ are not r.e. (unrecognizable). The matching option here is option D.