There are several vector/tensor calculus rules that will come in handy, so I will defined them here first (in no particular order):
$$
\begin{align}
\nabla \cdot \left[ \nabla \mathbf{A} - \left( \nabla \mathbf{A} \right)^{T} \right] & = \nabla \times \left( \nabla \times \mathbf{A} \right) \tag{0a} \\
\nabla \cdot \left( f \ \mathbf{A} \right) & = f \nabla \cdot \mathbf{A} + \nabla f \cdot \mathbf{A} \tag{0b} \\
\nabla \times \left( f \ \mathbf{A} \right) & = f \nabla \times \mathbf{A} + \nabla f \times \mathbf{A} \tag{0c} \\
\nabla \left( \mathbf{A} \cdot \mathbf{B} \right) & = \mathbf{A} \times \left( \nabla \times \mathbf{B} \right) + \mathbf{B} \times \left( \nabla \times \mathbf{A} \right) + \left( \mathbf{A} \cdot \nabla \right) \mathbf{B} + \left( \mathbf{B} \cdot \nabla \right) \mathbf{A} \tag{0d} \\
\mathbf{A} \cdot \left( \nabla \mathbf{B} \right)^{T} & = \left( \mathbf{A} \times \nabla \right) \times \mathbf{B} + \mathbf{A} \left( \nabla \cdot \mathbf{B} \right) \tag{0e} \\
\left( \nabla \mathbf{B} \right) \cdot \mathbf{A} & = \mathbf{A} \times \left( \nabla \times \mathbf{B} \right) + \left( \mathbf{A} \cdot \nabla \right) \mathbf{B} \tag{0f} \\
\nabla \cdot \left( \mathbf{A} \times \mathbf{B} \right) & = \mathbf{B} \cdot \left( \nabla \times \mathbf{A} \right) - \mathbf{A} \cdot \left( \nabla \times \mathbf{B} \right) \tag{0g} \\
\nabla \times \left( \nabla \times \mathbf{A} \right) & = \nabla \left( \nabla \cdot \mathbf{A} \right) - \nabla^{2} \mathbf{A} \tag{0h} \\
\nabla \times \left( \mathbf{A} \times \mathbf{B} \right) & = \nabla \cdot \left( \mathbf{A} \mathbf{B} \right)^{T} - \left( \mathbf{A} \mathbf{B} \right) \tag{0i}
\end{align}
$$
From these relations, one can show the following:
$$
\nabla \cdot \left\{ \eta \left[ \nabla \mathbf{B} - \left( \nabla \mathbf{B} \right)^{T} \right] \right\} = \nabla \times \left( \eta \nabla \times \mathbf{B} \right) + \left( \nabla \eta \cdot \nabla \right) \mathbf{B} - \left( \nabla \eta \times \nabla \right) \times \mathbf{B} \tag{1}
$$
where we have taken advantage of Maxwell's equations to eliminate the divergence of the magnetic field term.
The second term on the right-hand side can be expanded to the following form:
$$
\left( \nabla \eta \cdot \nabla \right) \mathbf{B} = \left( \mathbf{B} \cdot \nabla \right) \nabla \eta - \nabla^{2} \eta \mathbf{B} - \nabla \times \left( \nabla \eta \times \mathbf{B} \right) \tag{2}
$$
where all the terms on the right-hand side involve second order derivatives of $\eta$.
Generally, to simplify this down to make the two expressions of interest in your question equal one needs to make assumptions about the properties of the system. For instance, the $\nabla \times \left( \eta \nabla \times \mathbf{B} \right)$ comes from an approximation of Ohm's law and Ampere's law, i.e., $\mathbf{E} \approx \eta \mathbf{j}$ and $\mathbf{j} \propto \nabla \times \mathbf{B}$. If there are no local electric field sources (i.e., no excess charges), then $\nabla \cdot \mathbf{E} = 0$, which implies:
$$
\nabla \cdot \left( \eta \mathbf{j} \right) = \eta \nabla \cdot \mathbf{j} + \nabla \eta \cdot \mathbf{j} = 0 \tag{3}
$$
If $\mathbf{j} \propto \nabla \times \mathbf{B}$ is true, then the first term is zero as the divergence of the curl of a vector is always zero so we are left with:
$$
\nabla \cdot \left( \eta \mathbf{j} \right) \approx \nabla \eta \cdot \mathbf{j} = 0 \tag{4}
$$
The right-hand side can be rewritten as $\nabla \eta \cdot \left( \nabla \times \mathbf{B} \right) = 0$. We can then use Equation 0g above to show that the following is also true:
$$
\nabla \cdot \left( \nabla \eta \times \mathbf{B} \right) = 0 \tag{5}
$$
where we have used the fact that the curl of the gradient of a scalar is always zero.
We also know another relationship from Faraday's law where:
$$
\begin{align}
\nabla \times \mathbf{E} & = - \frac{ \partial \mathbf{B} }{ \partial t } \tag{6a} \\
& = \nabla \times \left( \eta \ \mathbf{j} \right) \tag{6b} \\
& = \eta \nabla \times \mathbf{j} + \nabla \eta \times \mathbf{j} \tag{6c} \\
& = \frac{ 1 }{ \mu_{o} } \left[ \eta \nabla \times \left( \nabla \times \mathbf{B} \right) + \nabla \eta \times \left( \nabla \times \mathbf{B} \right) \right] \tag{6d} \\
& = \frac{ 1 }{ \mu_{o} } \nabla \times \left( \eta \nabla \times \mathbf{B} \right) \tag{6e}
\end{align}
$$
where $\mu_{o}$ is the permeability of free space.
My question is: are these two expressions equal?
In general, no. Under the right approximations, yes.
