The symbol :
$\Gamma \vDash S$
means that the sentence $S$ is a logical consequence of the set of sentences $\Gamma$.
This means, as said in the above comments :
every interpretation that satisfies all sentences $\varphi \in \Gamma$ (i.e. is a model of $\Gamma$), satisfies also $S$.
The symbol :
$\Gamma \vdash S$
means that the sentence $S$ is provable from the set of assumptions $\Gamma$.
This means that the problem implicitly assume that we have in place a proof system, made of (one or more) rules of inference and (zero o more) axioms and a definition of derivation; in this way, the symbol above is defined (as in the comment) as :
there is a finite set of sentences $\sigma_1, … ,\sigma_n$, where either: (i) $\sigma_i \in \Gamma$, or (ii) $\sigma_i$ is a logical axiom or (iii) $\sigma_i$ follows by a rule of inference from some (one or more) $\sigma_k$, where $1 \le k < i$ and $\sigma_n = S$.
Finally, the symbol :
$\Gamma \nvdash P \land \lnot P$
means that form $\Gamma$, with the proof system above, we can derive a contradiction, i.e. $\Gamma$ is inconsistent. So, we say that $\Gamma$ is consistent when it is not inconsistent (i.e.when we cannot derive a contradiction from it).
The statement of the problem boils down to :
show that if $\Gamma$ is consistent, then it is satisfiable.
Now, in order to prove it, procede as suggested by @Ryan Sullivant, i.e. by contradiction, and suppose that $\Gamma$ is not satisfiable. This means that $\Gamma$ has no models.
Now re-read the relation of logical consequence explained above : $\Gamma \vDash \varphi$ iff every model of $\Gamma$ is also a model of $\varphi$.
But we assumed that $\Gamma$ has no models; so the last condition is vacuosly satisfied for every $\varphi$, and also for $P \land \lnot P$.
This means that, if $\Gamma$ is not satisfiable, i.e. if it has no models, than :
$\Gamma \vDash P \land \lnot P$.
At this point, we need the theorem mentioned in the hypotheses of the problem :
if $\Gamma \vDash S$, then $\Gamma \vdash S$
(if $S$ is a logical conseuqnce of $\Gamma$, then $S$ is provable from $\Gamma$; we say that the proof system is complete).
Having proved, form the assumption that $\Gamma$ is unstaisfiable, that $P \land \lnot P$ is a logical consequence of $\Gamma$, we use the above theorem to conclude that $P \land \lnot P$ is provable form $\Gamma$.
In conclusion :
if $\Gamma$ is unsatisfiable, then $\Gamma$ is inconsistent.