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.