= Solution
Put $x=\mu_{\rm B}B/(k_{\rm B}T)$. Summing the two <Boltzmann factors> gives the occupied-site internal <partition function>
$$
\boxed{z=e^{-\varepsilon_+/k_{\rm B}T}+e^{-\varepsilon_-/k_{\rm B}T}
=2e^{E_b/k_{\rm B}T}\cosh x.}
$$
If the single atom can occupy any of the $M$ identical sites, its complete surface <partition function> is
$$
\boxed{q_{\rm ads}=Mz=2M e^{E_b/k_{\rm B}T}\cosh x.}
$$
One also commonly calls the one-site factor $z$ itself $q_{\rm ads}$. Both conventions are valid, but the combinatorial factor in part (c) must use the same convention. We use the complete one-atom surface function $q_{\rm ads}=Mz$ below and give the equivalent one-site expression.
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