Let $X$ be a metric space, and $X'$ be a completion of $X$. Is $X'=\bar X$?
For any metric space $M$, one can construct a complete metric space $M′$ (which is also denoted as $\bar M$), which contains $M$ as a dense subspace.(Wikipedia)
What does it mean to contain $M$ as a dense subset? $M'=:\bar M$?
Let $(X,d)$ and $(Y, \widetilde d)$ be two metric spaces such that $A\subset X\cap Y$. Then you should note that (in general) the closure of $A$ in $X$ (denoted by $\bar A^X$) and the closure of $A$ in $Y$ (denoted by $\bar A^Y$) need not be the same.
$M'$ contains $M$ as a dense set means that $\bar M^{M'}=M'$ (which I think is what you mean). So the closure of $M$ in $M'$ is the whole space $M'$. Of course, we always have $\bar M^M=M$ since $M$ is always closed in itself.
Remark. In the Wikipedia article that you mentioned, $M$ is contained in $M'$ up to an isomorphism, similarly to how the rational numbers are contained in the construction of the real numbers.