From this we find the probability of a hidden unit being "on" or "off":
$$ \begin{align*} p_{BB} (h_j=1 | \boldsymbol{x}) =& \frac{ e^{(b_j + \boldsymbol{x}^T \boldsymbol{w}_{\ast j} ) h_j} } {1 + e^{b_j + \boldsymbol{x}^T \boldsymbol{w}_{\ast j}}} \\ =& \frac{ e^{(b_j + \boldsymbol{x}^T \boldsymbol{w}_{\ast j} )} } {1 + e^{b_j + \boldsymbol{x}^T \boldsymbol{w}_{\ast j}}} \\ =& \frac{ 1 }{1 + e^{-(b_j + \boldsymbol{x}^T \boldsymbol{w}_{\ast j})} } , \end{align*} $$and
$$ \begin{align*} p_{BB} (h_j=0 | \boldsymbol{x}) =\frac{ 1 }{1 + e^{b_j + \boldsymbol{x}^T \boldsymbol{w}_{\ast j}} } . \end{align*} $$