= Solution
For the adsorbed <lattice gas>, the <Helmholtz free energy> is
$$
A_{\rm ads}=-k_{\rm B}T\left[\ln\binom MN+N\ln z\right].
$$
Use <Stirling's approximation> with $M,N,M-N$ large, then differentiate at fixed $M,T,B$. Writing <surface coverage> as $\theta=N/M$ gives the <adsorption chemical potential>
$$
\boxed{\mu_{\rm ads}=k_{\rm B}T\ln\frac{\theta}{1-\theta}
-k_{\rm B}T\ln z
=-E_b-k_{\rm B}T\ln(2\cosh x)
+k_{\rm B}T\ln\frac{\theta}{1-\theta}.}
$$
The occupation ratio comes from site-mixing <entropy>; it is essential near saturation.
Let $N_g$ be the number of free atoms and $n_g=N_g/V$. In the field-free gas, both <spin> states have zero <energy>, so the one-atom <partition function> is $q_g=2V/\Lambda^3$, where $\Lambda$ is the <thermal de Broglie wavelength>. For a dilute <ideal gas>,
$$
Q_g=\frac{(2V/\Lambda^3)^{N_g}}{N_g!},\qquad
A_g\simeq N_gk_{\rm B}T\left[\ln\frac{n_g\Lambda^3}{2}-1\right].
$$
Therefore
$$
\boxed{\mu_g=k_{\rm B}T\ln\frac{n_g\Lambda^3}{2}
=k_{\rm B}T\ln\frac{p\Lambda^3}{2k_{\rm B}T}.}
$$
The last equality uses $p=n_gk_{\rm B}T$. The bulk <spin degeneracy> is two even though the bulk <magnetic field> vanishes: degeneracy is not the absence of <spin> states.
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