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Let $\kappa \geq \aleph_0$. Let $A \subseteq M$ with $|A|< \kappa$. Let $\mathcal{M}_A$ be the $L_A$- structure obtained from $\mathcal{M}$ by interpreting the new constant symbols in the natural way. If M is $\kappa$-saturated, then so is $\mathcal{M}_A$.

Would be grateful for your hints, or solutions.

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    Let $M$ and $A$ and $\kappa$ be as given. If you take a set $B'\subseteq{M_A}$ of size less than $\kappa$, then $A\cup{B}$ is a subset of size less $\kappa$ of $M$. Further, any type over $B$ in $M_A$ is type over $A\cup{B}$ in $M$ and thus is realized in $M$ and hence $M_A$. – 2017-01-17
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    @dav11 I think answers (even easy ones!) should be posted as answers, not comments. – 2017-01-17
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    @AlexKruckman: Wasn't sure if this was worth posting as an answer. But I will post it as an answer. – 2017-01-17

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As for Alex's suggestion, I'm posting the comment as a answer:

Let $M$ and $A$ and $\kappa$ be as given. If you take a set $B\subseteq{M_A}$ of size less than $\kappa$, then $A\cup{B}$ is a subset of size less than $\kappa$ of $M$. Now you can show that any consistent type over $B$ in $M_A$ is a consistent type over $A\cup{B}$ in $M$ and thus is realized in $M$ and hence $M_A$.