Consider a set $\Gamma$ of formula which is satisfiable in language $\mathcal{L}$. let $\mathcal{L'} \supseteq \mathcal{L}$ be any expansion. then is $\Gamma$ satisfiable in $\mathcal{L'}$?
I can consider when $C$ a set of new constants are added(by completeness)
Yes it's still satisfiable. Suppose $M$ is some structure such that $M\models \Gamma$. Expand $M$ to a structure $M'$: same underlying set, with arbitrary and/or trivial interpretations of the nonlogical constants of $\cal{L'}$ that are not in $\cal{L}$ — empty predicates (for example), arbitrary values for constants and functions. Then it's easy to see (yes?) that $M'\models \Gamma$ too, as the new symbols don't occur in $\Gamma$ and their interpretations are never used in evaluating its sentences.
Your last sentence presumably means that $\cal{L'}$ might arise from $\cal{L}$ by adding constants, as in Henkin's completeness proof for first order logic. I think so, but I'm not entirely sure.