A similar derivation yields the marginal probability of the hidden units
$$ \begin{align*} p_{BB} (\boldsymbol{h},\boldsymbol{\Theta}) = \frac{1}{Z_{BB}(\boldsymbol{\Theta})} e^{\boldsymbol{b}^T \boldsymbol{h}} \prod_i^M (1 + e^{a_i + \boldsymbol{w}_{i\ast}^T \boldsymbol{h}}) . \end{align*} $$