$\def \NP {\rm NP}$ $\def \NPC {\rm NPC}$ $\def \co {\rm co \text -}$
(a) NP
Definition of $\co\NP$: If a language $L \in \NP$, then its complement $L' \in \co \NP$
Consider an arbitrary language $M \in \NP$. Thus, $M' \in \co\NP$.
Since $L \in \NPC$, there is a polynomial time reduction from $M$ to $L$.
This means that there is a polynomial time reduction $R(x)$ such that $x \in M$ if and only if $R(x) \in L$. $$x \in M \Longleftrightarrow R(x) \in L$$
Also, $x \in M'$ if and only if $R(x) \in L'$. $$x \in M' \Longleftrightarrow R(x) \in L'$$
This usually only allows us to decide $M$ in $\NP$ time. To decide if $x \in M$, we can simply compute $R(x)$ in polynomial time, and check if $R(x) \in L$ in $\NP$ time.
If we try to decide $M'$, we need to solve $L'$ which is $\co\NP$, so we might not be able to solve it in $\NP$ time. (We don't know if $\NP = \co\NP$)
However, if our $L$ is such that $L \in \NPC$ and $L' \in \NP$, then we can also decide $M'$ in $\NP$ time.
Thus, no matter what problem $M \in \NP$ we have, we can decide both $M$ and $M'$ in $\NP$ time. Which means $\NP = \co\NP$
What about option (b)?
Even if the complexity classes $\NP$ and $\co\NP$ collapse into one, the $\rm P$ vs $\NP$ question remains open. Our assumed language $L$ doesn't imply option (b).
