If $\text{P}(M|D)$ is posterior, $a$ is the proportionality constant, $\text{P}(M)$ is the prior and $\text{P}(D|M)$ is the likelihood. I have the the prior distribution, and I know the function that can give the likelihood for any input parameter sample. Then the Bayes rule is: $\text{P}(M|D)= a\times\text{P}(M)\times\text{P}(D|M)$.
If I want to sample from $\text{P}(M|D)$ by Gibbs, what is my full conditional distribution? is it $\text{P}(M)\times\text{P}(D|M)$?
No, $p(M)p(D\vert M)$ is the product of the likelihood function and the prior, and $a$ is the normalising constant, explained by the (take off your hats please) Bayes' Theorem.
If you want the full conditionals of your model in order to construct a Gibbs sampler, then you need to provide more information about your model. Only few distributional assumptions/ prior choices lead to closed-form conditionals. (See the wikipedia article for a precise description of of the "conditional distributions" required to construct a Gibbs sampler).
In the worst scenario where you cannot obtain closed-form conditionals, you can still construct a Metropolis-within-Gibbs sampler.
