If $A$ is unital C$^*$algebra, is it true that the multiplier algebra of $A \otimes \mathcal{K} $ is $ A \otimes \mathcal{B}(\mathcal{H})$? Where $\mathcal{K}$ is C$^*$algebra of compact operators on the Hilbert space $\mathcal{H}$.
2 Answers
The fact stated in the answer by vap is proven in the paper "Multipliers of C*algebras" by Akemann, Pedersen and Tomiyama (see Theorem 3.3, I guess). Moreover, they prove in Theorem 3.8 that multiplier algebras are not very well behaved with respect to minimal tensor products:
Let $A$ and $B$ be $C^*$algebras and assume that $B$ has a countable approximate unit, but no unit (think of $\mathcal{K}$ here in your case) and that $A$ is infinite dimensional. Then
$$ M(A) \otimes M(B) \subsetneq M(A \otimes B) $$ where the tensor product is the minimal one.
So, in particular, for any infinite dimensional unital $C^*$algebra $A$, the tensor product $A \otimes \mathcal{B}(\mathcal{H})$ is always a proper subalgebra of $M(A \otimes \mathcal{K})$.
If $A=C_0(X)$ and $B$ is a $C^\ast$algebra then $M(A\otimes B)$ is the set of strictly continuous functions $\beta X\to M(B)$, where $\beta$ stands for StoneČech compactification.
If you take $X$ to be compact and $B=\mathcal{K}$ then we are in your setting.
But $C(X)\otimes\mathcal{B(H)}$ is the set of normcontinuos functions $\beta X=X\to\mathcal{B(H)}$. The strict topology is the $\sigma$strong$^\ast$ topology on $\mathcal{B(H)}$, which is different form the norm topology. This should answer your question in the negative.

$\begingroup$ What space $X$ do you have in mind? (It must be infinite, of course.) $\endgroup$ May 25, 2016 at 15:06

3$\begingroup$ The onepoint compactification of the natural numbers should work. We can use this space to index a sequence converging in the strict topology of bounded operators, but not converging in norm. $\endgroup$– vapMay 25, 2016 at 16:15

1$\begingroup$ For future reference, as pointed out in the comments of mathoverflow.net/questions/375906 this is not true. $M(A\otimes B)$ may be identified with the space of strictly continuous, bounded functions $X\rightarrow M(B)$, and not from $\beta X$. $\endgroup$ Nov 8, 2020 at 12:13

1$\begingroup$ @MatthewDaws Thanks. The error probably stems from the universal property of the StoneČech compactification: one would like to upgrade bounded functions on $X$ to continuous functions on the compactification. But this only works if the bounded sets in $M(B)$ are compact in the strict topology, which I think can fail even when $B$ is commutative. $\endgroup$– vapNov 9, 2020 at 13:08