Can exactly one of NP and co-NP be equal to P?
No, because $\mathsf{P}$ is closed to complement this cant be, and we even know that $\mathsf{P}=\mathsf{NP} \implies \mathsf{NP}=\mathsf{co\text{-}NP}$.

Let us assume that $\mathsf{P}=\mathsf{NP}$, and let $L \in \mathsf{co\text{-}NP}$, thus $L^c \in \mathsf{NP}$. We assumed $\mathsf{P}=\mathsf{NP}$, and therefore there exists a TM $M$ s.t. $L(M)=L^c$. If we take the "complement" of $M$, that is a machine $M'$ which is identical to $M$ except its accept and reject states are reversed, we get that $L(M')=(L^c)^c=L$, and thus $L$ is in $\mathsf{NP}$.

This shows that, under the assumption that $\mathsf{P}=\mathsf{NP}$, we get $(\mathsf{P}=)\mathsf{NP}=\mathsf{co\text{-}NP}$ and thus $\mathsf{P}=\mathsf{co\text{-}NP}$.