= Solution
Apply $\widehat{\mathbf z}\mathbin\cdot\nabla\times$ to the linear momentum equation. The pressure and buoyancy terms vanish, while incompressibility gives $\widehat{\mathbf z}\cdot\nabla\times(\widehat{\mathbf z}\times\mathbf u')=-D W$. Hence
$$
\boxed{(\partial_t-\sigma\nabla^2)\omega-\lambda DW=0}.
$$
Applying $\widehat{\mathbf z}\cdot\nabla\times\nabla\times$ and using the identity supplied in the question gives
$$
\boxed{
(\partial_t-\sigma\nabla^2)\nabla^2W
+\lambda D\omega
=\sigma Ra\,\nabla_H^2\theta'}.
$$
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