Fix a language $L$, a model $M$ and a set $A\subseteq M$. Let $p(x,y)\subseteq L(A)$ be a type. If $M$ is saturated and $|A|<|M|$ then for every $a\in M$
$$M\models \exists y\ p(a,y)\leftrightarrow \bigwedge\Big\{\exists y\ \varphi(a,y)\ :\ \varphi(x,y)\in p\Big\}.$$
(In case notation is not self explanatory: in the RHS the witness of $\exists y$ depends on $\varphi$, while on the LHS it does not.)
Do you have an example that proves that saturation is necessary?
Let $p(y)$ be absolutely any consistent type over any set $A\subseteq M$ which is not realized in $M$. Then you have an example: letting $x$ be the empty tuple of variables, $M\not\models \exists y\, p(y)$, but for all $\varphi(y)\in p$, $M\models \exists y\,\varphi(y)$, since $p$ is consistent. So saturation is precisely what is necessary and sufficient for the claim to be true for all $p$ over all $A$ with $|A|< |M|$.
But maybe this is too trivial. For a concrete example in which $x$ is not empty, let $T = \text{Th}(\mathbb{Z},<)$ (the theory of discrete linear orders without endpoints), and let $p(x,y)$ be the type-definable relation "$x$ is infinitely far below $y$": $\{\exists^{\geq n} z\, (x < z < y)\mid n\in\omega\}$. Then $\mathbb{Z}\not\models \exists y\, p(0,y)$, but for all $n$, $\mathbb{Z}\models \exists y \exists^{\geq n} z\, (x < z < y)$.