Is there an accessible proof for the following fact?
If $A=C_0(X)$ with $X$ locally compact Hausdorff and $B$ is a $C^\ast$-algebra then $M(A\otimes B)$ is the set of bounded strictly continuous functions $X \to M(B)$.
Denote the set of bounded strictly continuous functions by $C_b^s (X, M(B))$.
Thanks to the hint in the comments, we can say the following:
Given $x \in X$, there is a mapping $$\pi_x: C_0(X) \otimes B \to B: f \otimes b \mapsto f(x)b$$ which extends to a map $$\pi_x: M(C_0(X) \otimes B) \to M(B)$$ and this allows us to define $$M(C_0(X) \otimes B) \to C_b^s(X,M(B)): L \mapsto (x \mapsto \pi_x(L))$$
Why is this an isomorphism of $C^*$-algebras, i.e. why is it injective and surjective?