If $A$ is a $C^*$ algebra, I know the fact: If $\tau$ is a state on $A$, we can extend it to a state on $M(A)$, by continuity of $\tau$ and where $M(A)$ is the multiplier algebra of $A$.
If $\tau$ is a tracial state on $A$, can we extend $\tau$ to get a tracial state on $M(A)$?
Let $\tau$ be a state on $M(A)$ such that its restriction to $A$ satisfies $\tau(ab) = \tau(ba)$ for all $a,b \in A$. For $x,y \in M(A)$ we then have $$ \tau(xy) = s-\lim\tau((xe_\lambda)(e_\lambda y)) = s-\lim \tau(e_\lambda yx e_\lambda) = \tau(yx), $$ where $(e_\lambda)_\lambda$ is an approximate unit for $A$. Note that $e_\lambda \to 1$ strictly and hence $e_\lambda x \to x$ strictly. Since $\tau$ is strictly continuous, the result follows.
Note that in the case $A$ is separable, $A$ is a hereditary C*-subalgebra of $M(A)$, namely $A = \overline{hM(A)h}$, where $h \in A$ is strictly positive. In that case every state on $A$ has a unique extension to $M(A)$ given by $$ \tau(x) = \lim \tau(e_\lambda x e_\lambda) \qquad (x \in M(A)). $$
Edit: For the argumet to work one needs that $\tau$ is strictly continuous on $M(A)$. I will check if this is automatic and then update the answer.