In order to find the probability of any configuration of the visible units we derive the marginal probability density function.
$$ \begin{align*} p_{BB} (\boldsymbol{x},\boldsymbol{\Theta}) =& \sum_{\boldsymbol{h}} p_{BB} (\boldsymbol{x}, \boldsymbol{h},\boldsymbol{\Theta}) \\ =& \frac{1}{Z_{BB}} \sum_{\boldsymbol{h}} e^{\boldsymbol{a}^T \boldsymbol{x} + \boldsymbol{b}^T \boldsymbol{h} + \boldsymbol{x}^T \boldsymbol{W} \boldsymbol{h}} \nonumber \\ =& \frac{1}{Z_{BB}} e^{\boldsymbol{a}^T \boldsymbol{x}} \sum_{\boldsymbol{h}} e^{\sum_j^N (b_j + \boldsymbol{x}^T \boldsymbol{w}_{\ast j})h_j} \nonumber \\ =& \frac{1}{Z_{BB}} e^{\boldsymbol{a}^T \boldsymbol{x}} \sum_{\boldsymbol{h}} \prod_j^N e^{ (b_j + \boldsymbol{x}^T \boldsymbol{w}_{\ast j})h_j} \nonumber \\ =& \frac{1}{Z_{BB}} e^{\boldsymbol{a}^T \boldsymbol{x}} \bigg ( \sum_{h_1} e^{(b_1 + \boldsymbol{x}^T \boldsymbol{w}_{\ast 1})h_1} \times \sum_{h_2} e^{(b_2 + \boldsymbol{x}^T \boldsymbol{w}_{\ast 2})h_2} \times \nonumber \\ & ... \times \sum_{h_2} e^{(b_N + \boldsymbol{x}^T \boldsymbol{w}_{\ast N})h_N} \bigg ) \nonumber \\ =& \frac{1}{Z_{BB}} e^{\boldsymbol{a}^T \boldsymbol{x}} \prod_j^N \sum_{h_j} e^{(b_j + \boldsymbol{x}^T \boldsymbol{w}_{\ast j}) h_j} \nonumber \\ =& \frac{1}{Z_{BB}} e^{\boldsymbol{a}^T \boldsymbol{x}} \prod_j^N (1 + e^{b_j + \boldsymbol{x}^T \boldsymbol{w}_{\ast j}}) . \end{align*} $$