Identifying the multiplier $C^*$-algebra $M(C_0(X) \otimes B)$ 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?
 A: Indeed, Blackadar appears to have made a mistake here.  But the reference he gives is good, and seems to give both a correct statement, and a proof:
Akemann, Charles A.; Pedersen, Gert K.; Tomiyama, Jun
Multipliers of C∗-algebras.
J. Functional Analysis 13 (1973), 277–301.
MR470685 DOI: 10.1016/0022-1236(73)90036-0
See Corollary 3.4.
Edit: How I would approach this.  Firstly, understand carefully the proof that $C_0(X) \otimes B \cong C_0(X,B)$.  Much the same ideas are used for the multiplier algebra case.  In particular, for $f\in C_0(X), b\in B$ we identify $f\otimes b$ with the continuous map $X\rightarrow B; x\mapsto f(x) b$.
I would look at $\Phi: C^b_{str}(X, M(B)) \rightarrow M(C_0(X,B))$ defined by pointwise multiplication:

*

*First show this is well-defined.  This is easy, as for $F\in C^b_{str}(X, M(B))$ we have that $F(f\otimes b) \in C_0(X,B)$.  Then copy the proof that $C_0(X) \otimes B \cong C_0(X,B)$ to show that $F$ does multiply $C_0(X,B)$ into itself.

*$\Phi$ is clearly injective.

*To show $\Phi$ is surjective, argue as in the OP: given $L\in M(C_0(X,B))$ we define $F(x) = \pi_x(L)$.  Then $X\rightarrow B; x \mapsto \pi_x(L) f(x) b
= \pi_x(L(f\otimes b))$ is continuous, for each $f\in C_0(X)$ and $b\in B$.  This is enough to show that $F$ is continuous, for the strict topology

*Checking that $\Phi$ is a $*$-homomorphism is routine.

