The answer is indeed yes. For this let us consider the definition of the truth value of $\forall x:\varphi(x)$. Such sentence is true if and only if for every $x\in M$ the formula $\varphi$ is true for $x$.
This may seem circular, however we look at the structure $M$ and if we can tell externally that every $x$ satisfies $\varphi$, then $M$ satisfies $\forall x:\varphi(x)$.
Now suppose that in $M$ the following is true $\forall x: (p(x)\land q(x))$.
That to say that for every $x\in M$ the sentence $p(x)\land q(x)$ is true in $M$.
Therefore for every $x$ (in $M$) we have that $p(x)$ holds and $q(x)$ holds (simply because $p(x)\land q(x)$ holds if and only if $p(x)$ and $q(x)$ hold).
So for every $x\in M$ we have that $p(x)$, thus $\forall x:p(x)$ and for all $x\in M$ we have $q(x)$, so $\forall x:q(x)$.
Hence, the conjunction of these statement is true.
Conversely, suppose in $M$ it is true that $\forall x:p(x)\land \forall x:q(x)$.
For every $x\in M$ then $p(x)$ and $q(x)$ both hold, from the assumption.
So for every $x\in M$ holds $p(x)\land q(x)$.
Thus, in $M$ it is true that $\forall x : (p(x)\land q(x))$.