My approachLet L  P$\exists$ Turing Machine $M_1$ which decides L. We can easily construct $M_2$ which decides $\bar{L}$$\bar{L}$  CO-NP $\implies$ P  Co-NPI'm not sure whether its a correct way to prove this or not. I found this method here Link
How to prove P  Co-NP
complexity theory;complexity classes;np;p vs np
It's because P = Co-P: if we can find a solution for a problem in Polynomial time, we can also decide in Polynomial time that it does NOT have a solution. Now, since P  NP we have that the complement of this problem is in Co-NP. But the complement of any problem in P is in Co-P and therefore any complement of any P problem is in Co-NP. And since the complement of any P problem is Co-P = P, we have P  Co-NP