Let $X$ be a Banach space and $M$ be a von Neumann algebra, and consider $B(X, M)$, the space of bounded linear operators between $X$ and $M$. Let $E_0 \subset$ $B(X, M)^*$ be the space $$ E_0=\operatorname{Span}\left\{x \otimes \xi \in B(X, M)^*: x \in X, \xi \in M_*\right\}, $$ where $M_*$ is known as the predual of $M$, consisting of all normal linear functionals on $M$, and we define $(x \otimes \xi)(T)=\xi(T x)$ for all $T \in B(X, M)$.
Let $E=\overline{E_0}^{\|\cdot\|}$, then $E$ is a Banach space. The problem is to verify that $E^* \cong B(X, M)$ (isometric isomorphism) via the map $$i:B(X,M) \longrightarrow E^* \\ T \mapsto [\varphi \mapsto \varphi(T)].$$
Unfortunately, I got stuck here. I guess the process would be first to verify that the map is an isometry (hence injective) and then verify that this is also subjective. However, have no idea to do both.
Any help is appreciated.
I think this one is standard, probably around in Paulsen's book or Brown & Ozawa. I believe it's called the "bounded weak" topology. For each $T$, your functional $\phi_T: E \to \mathbb{C}$ is given by $\phi_T(x \otimes y) = x \otimes y(T)$ (which is $T(x)(y)$) on elementary tensors, and so your map $i$ is just $i(T) = \phi_T$ (check that $\|\phi_T\| \leq \|T\|$, so that $\phi_T$ can actually be extended to the whole of $E$, and that $\phi_T \in E^*$). So now the question becomes whether or not $i: B(X,(M_*)^*) \to E^*$ given by $i(T) = \phi_T$ is a surjective isometry.
Checking these can be done concurrently. Take a linear functional $\zeta \in E^*$ and define a functional $\zeta_x$ on $M_*$ by $\zeta_x(\xi) = \zeta(x \otimes \xi)$. Convince yourself that $\zeta_x$ is bounded, so that we do really have $\zeta_x \in (M_*)^*$ for all $x$. Now define $T: X \to (M_*)^*$ by $T(x) = \zeta_x$, which is linear and satisfies $\|T\| \leq \|\zeta\|$ (check this). Moreover, $\Phi_T = \zeta$ since they are both continuous and equal on the dense subset which is spanned by pure tensors (and consequently $\|\Phi_T\| = \|\zeta\|$ as well). This gives surjectivity. Moreover, we also have $\|T\| \leq \|\zeta\| = \|\phi_T\| \leq \|T\|$, giving the isometry condition.
Just a comment that you can actually get an appropriate surjective isometry $$ \overline{\text{span}}\{x \otimes y \in B(X,Y^*)^* \mid x \in X, y \in Y\} \simeq B(X,Y^*), $$ whenever $Y$ is a Banach space (replace $M_*$ with $Y$ in all of the above).