Lel $B$ be a Banach algebra and give $B^{\*\*}$ one of the Arens products in order to make it a Banach algebra. Then the canonical embedding $\kappa\colon B\to B^{\ast\ast}$ is a homomorphic embedding wrt to either Arens product. 


Suppose $M$ is a maximal abelian subalgebra of $B$. Can infer from this that $\overline{\kappa(M)}^{w^\*}$ is a maximal abelian subalgbra of $B^{\ast\ast}$? I suspect it need not be the case... How about when $B$ is Arens regular, that is, when the Arens products in $B^{\ast\ast}$ coincide? 



Masas in second duals of Banach algebras
It need not be so. The simplest example would be the Toesplitz algebra $\cal T$. It is the ${\rm C}^\ast$-algebra generated by the unilateral shift $S$ on $\ell\_2{\bf N}$, $Se\_n=e\_{n+1}$. The algebra ${\cal T}$ contains the ideal $\cal K$ of the compact operators, and the quotient is $\pi\colon{\cal T} \to {\cal T}/{\cal K} \cong C({\bf T})$. The diagonal subalgebra $D:=\ell\_\infty{\bf N} \cap {\cal T}$ is a masa of $\cal T$, which coincides with $c\_0{\bf N}+{\bf C}1$. This follows from the facts that $(c\_0{\bf N})'\cap{\bf B}(\ell\_2{\bf N})=\ell\_\infty{\bf N}$ and that $\langle xe\_n,e\_n\rangle$ converges for every $x\in{\cal T}$ (to $\int\_{\bf T} \pi(x)\ dm$). One has $$\bigl( D^{\ast\ast}\subset {\cal T}^{\ast\ast} \bigr) \cong \bigl( (c\_0{\bf N})^{\ast\ast}\oplus {\bf C}1\subset {\cal K}^{\ast\ast} \oplus C({\bf T})^{\ast\ast} \bigr)$$
and $D^{\ast\ast}$ is not a masa.


In fact, I think the weak\*-closure of a masa is never a masa in the second dual $B^{\ast\ast}$ unless the situation is very degenerate, e.g., $c\_0 \subset {\cal K}$ or the masa has finite index. This is because every quasi-equivalence class of a \*-representation $\pi$ of $B$ has a corresponding central projection $c([\pi])$ in $B^{\ast\ast}$, which satisfies $\pi(B)'' \cong B^{\ast\ast}c([\pi])$. That $\overline{M}^{w\ast}$ contains them all means that the quasi-equivalence class of a \*-representation $\pi$ of $B$ is uniquely determined by its restriction $\pi|\_M$ to $M$. This is rather rare.


