Is the open question NP=co-NP the same as P=NP?
No. It is another open problem and certainly related, but different. The complexity class co-$\mathsf{NP}$ is the set of languages whose complements are in $\mathsf{NP}$; that is, the set of decision problems for which a "no" answer has a deterministic polynomial-time verifier. So for example, the question "Is this SAT formula unsatisfiable?" If the answer is "no", then there is some satisfying assignment of the variables that proves this; that's the certificate for the verifier.

It is possible that $\mathsf{P} \neq \mathsf{NP}$, yet $\mathsf{NP} = $co-$\mathsf{NP}$.

But on the other hand, if $\mathsf{P} = \mathsf{NP}$, then $\mathsf{NP} = $co-$\mathsf{NP}$ for sure. This is because if a language is in $\mathsf{P}$, then its complement is also in $\mathsf{P}$, so if $\mathsf{P} = \mathsf{NP}$, then that goes for every language in $\mathsf{NP}$ as well.