= Solution
At <adsorption> equilibrium, $\mu_{\rm ads}=\mu_g$. The expressions from part (d) give
$$
\frac{\theta}{1-\theta}
=z\frac{p\Lambda^3}{2k_{\rm B}T}
=\frac{p\Lambda^3}{k_{\rm B}T}e^{E_b/k_{\rm B}T}\cosh x.
$$
Define
$$
\boxed{b(T,B)=\frac{\Lambda^3}{k_{\rm B}T}
e^{E_b/k_{\rm B}T}\cosh\!\left(\frac{\mu_{\rm B}B}{k_{\rm B}T}\right).}
$$
Solving the occupation ratio yields
$$
\boxed{\theta=\frac{bp}{1+bp}.}
$$
The coefficient $b$ has units of inverse <pressure>, so $bp$ is dimensionless. The two <spin> states contribute a factor two on both sides of the <adsorption> equilibrium and cancel; retaining the surface factor but dropping the bulk degeneracy would incorrectly double $b$.
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