When $H,K$ are finite dimensional, this is well explained in Pisier's book *Introduction to operator space theory* in the chapter on Haagerup tensor product, and your space is just the completely bounded maps on $B(K,H)$. When $H$ and $K$ are infinite dimensional, the argument works identically with bounded operators replaced by compact operators. To obtain a similar identification for bounded operators, my guess is that it is better to work with larger completions (extended Haagerup tensor product).

Let me expand a bit. When $H,K$ are finite dimensional, $B(K,H) = H_c \otimes_h K^*_{r}$ (where $H_c$ denotes $H$ with the column operator space structure, and $K^*_r$ denotes $K^*$ with the row Hilbert space structure, the dual of $K_c$). More generally, $H_c\otimes_h \cdot = H_c \otimes_{min} \cdot$ and $\cdot \otimes_{h} H_r = \cdot \otimes_{min} H_r$: when one tensorizes on the left (resp. right) by column (resp. row) operator spaces, Haagerup and minimal tensor product coincide. Taking the dual, we obtain $B(K,H)^* = H^*_r \otimes_h K_c$.

Combining all this with the associativity of the Haagerup tensor product, we obtain the natural identifications
$$ B(H) \otimes_h B(K) = H_c \otimes_h (H_r^* \otimes_h K_c)\otimes_h K^*_r = H_c \otimes_{min} K^*_r \otimes_{min} B(H,K)^* = B(H,K) \otimes_{min} (B(K,H))^* = CB(B(K,H),B(K,H)).$$
For example, when $H=K=\mathbf{C}^n$, we obtain
$$ M_n \otimes_h M_n = M_n(S^1_n) = CB(M_n,M_n).$$