let $A\subseteq B\subseteq\mathcal{M}$ (where $\subseteq$ is just the subset symbol). $q=\mathrm{tp}(a/B)$ is an heir of $p=\mathrm{tp}(a/A)$ iff for any $\varphi(x;y)\in\mathcal{L_A}$, $\varphi(x;b)\in q$ for some $b\in B$ implies $\varphi(x;a)\in p$ for some $a\in A$.
I'm trying to prove when $\mathcal{M}\subseteq B$ (possibly $B$ subset of some $\mathcal{N}>\mathcal{M}$ where $\mathcal{N}$ is $|B|^+$-saturated) and $p\in S(\mathcal{M})$, there exists $q\in S(B)$ an heir of $p$.
I wanted to find $q$ contradicting "there exists $\varphi(x;b)\in q$ with $\forall a\in\mathcal{M}, \varphi(x;a)\notin p$." So letting $p=\mathrm{tp}(c/ \mathcal{M})$, I set $\Sigma:=\{\varphi(x;b)\in\mathcal{L}_B\mid \mathcal{N}\models\varphi(c;b)\wedge \forall a\in\mathcal{M},\mathcal{N}\models\varphi(c;a) \}$. I could prove $\Sigma \cup p$ is consistent so I took a completion $q$ of this set in $\mathcal{L}_B$. I assumed $q$ is not an heir of $p$ but failed to derive a contradiction. I tried to construct another set but I couldn't find an appropriate one.
In the definition of heir, you forgot the condition that $p\subseteq q$.
Hint: You almost chose the right set $\Sigma$, but not quite. Instead, look a: $$\Sigma = p\cup \{\varphi(x;b)\mid \varphi(x;y)\in \mathcal{L}_A, b\in B,\text{and } \forall a\in \mathcal{M}, \varphi(x;a)\in p\}.$$
Showing this $\Sigma$ is consistent will allow you to conclude, since if $q$ is any consistent completion of $\Sigma$, and if we assume for contradiction that $\varphi(x;b)\in q$ but $\varphi(x;a)\notin p$ for all $a\in \mathcal{M}$, then $\lnot \varphi(x;a)\in p$ for all $a\in \mathcal{M}$, so $\lnot \varphi(x;b)\in \Sigma\subseteq q$, contradicting consistency of $q$.
This argument won't work with your choice of $\Sigma$, since we don't necessarily have $\mathcal{N}\models \lnot \varphi(c;b)$.