Artem Chenikov and Itay Kaplan's paper 'Forking and Dividing in NTP2 Theories' remark 2.16 claims that every model is an extension base with respect to finite satisfiability:
For reference they define an extension base as follows:
Can anyone explain how the result "follows from the fact that filters can be extended to ultrafilters"?
For any model $M$ and any $a$ we have that $\operatorname{tp}(a/M)$ is finitely satisfiable in $M$. Namely, let $\varphi(x) \in \operatorname{tp}(a/M)$ (it is enough to check for single formulas, as types are closed under conjunction), then $\models \exists x \varphi(x)$ and so because $\exists x \varphi(x)$ only contains parameters in $M$ and $M$ is an elementary substructure of the monster model, we indeed get $M \models \exists x \varphi(x)$. So the problem reduces to showing that any type $p(x) = \operatorname{tp}(a/Cb)$ that is finitely satisfiable in some $C$ can be extended to a finitely satisfiable (in $C$) type over any parameter set $D \supseteq Cb$. One way to do this is by using that filters can be extended to ultrafilters as follows.
Consider the Boolean algebra $\mathcal{B}$ of formulas over $D$ in free variables $x$. Let $$ F_0 = \{ \varphi(x) \in \mathcal{B} : \varphi(C) = C \}. $$ and let $F$ be the filter generated by $p(x) \cup F_0$. We claim that $F$ is non-trivial (i.e. $F \subsetneq \mathcal{B}$). Clearly $F_0$ and $p(x)$ are closed under conjunctions, so it is enough to verify for $\psi(x) \in p(x)$ and $\varphi(x) \in F_0$ that $\psi(x) \wedge \phi(x)$ is consistent. Indeed, since $p(x)$ is finitely satisfiable in $C$ there is $c \in C$ such that $\models \psi(c)$ and hence $\models \psi(c) \wedge \phi(c)$. Now let $U$ be any ultrafilter on $\mathcal{B}$ extending $F$. Then $U$ is a complete type over $D$. We verify that $U$ is finitely satisfiable in $C$. Let $\varphi(x) \in U$. If $\varphi(C) = \emptyset$ then $\neg \phi(C) = C$, so $\neg \varphi(x) \in F_0 \subseteq F \subseteq U$. This contradicts $\varphi(x) \in U$, so we conclude that there must be $c \in C$ with $\models \varphi(c)$, as required.