 Solution

Add new constants $\bar{c}$ and considerEvery finite subset is satisfiable because $p$ is consistent over $M$. By the compactness theorem it has a model $N^{\prime }$. The constants naming $M$ give an elementary embedding $M\to N^{\prime }$, and ${\bar{c}}^{N^{\prime }}$ realizes $p$. Replacing $N^{\prime }$ by an isomorphic copy containing $M$ gives the required realization of a type in an elementary extension.