Is it possible that Co-NP = P but NP != P
The answer is no, and the reason is that $\mathrm{P}$ is trivially closed under complement. Just by definition of $\mathrm{co}$, it follows that $\mathrm{coNP} = \mathrm{P}$ is equivalent to $\mathrm{NP} = \mathrm{coP}$, and then we use that $\mathrm{coP} = \mathrm{P}$ to conclude that $\mathrm{coNP} = \mathrm{P}$ iff $\mathrm{NP} = \mathrm{P}$.