ANSWER: A,C
$\overline{L(N)}$ means Complement of the language accepted by NFA $N$. This is language complement.
$L(\overline{N})$ means Language accepted by the complement of NFA $N$. This is machine complement.
Given NFA $N$:-
First remove all non-reachable states:-
So $L(N) = \{ϵ\}$ clearly.
Thus $\overline{L(N)}$ which is the Complement of the language accepted by NFA $N = ∑^*-\{ϵ\}$.
Now, $\overline{N}$ can be drawn by changing all final to nonfinal states and vice-versa without touching the arrows,
the NFA $(\overline{N})$:-
Thus, $L(\overline{N})$ which is the Language accepted by the complement of NFA $N$ is also $ = \{ ϵ\}$. This is because the starting state itself is the final state. It would have been $\phi$ if there was no final state.
So, we have $L(N) = \{ϵ\}$, $\overline{L(N)}= ∑^*-\{ϵ\}$, $L(\overline{N})= \{ϵ\}$
Option A) $L(N) = L(\overline{N})= \{ϵ\}$ TRUE
Option B) $\overline{L(N)}$ intersection $L(\overline{N}) = ∑^*-\{ϵ\}$ intersection $\{ϵ\} = \phi$. FALSE
Option C) $\overline{L(N)} \cup L(\overline{N}) = ∑^*-\{ϵ\} \cup \{ϵ\} = ∑^*$ . TRUE