Multiplier algebra of $A \otimes \mathcal{K}$ 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 $\mathcal{H}$.
 A: 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 Stone-Čech 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 norm-continuos 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.
A: 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 $M(A \otimes \mathcal{K})$.
