= Solution
At fixed <temperature>,
$$
\boxed{b(B)=b(0)\cosh x,\qquad x=\mu_{\rm B}B/(k_{\rm B}T).}
$$
Thus <magnetic-field-dependent Langmuir adsorption> is even in $B$ and increases with $|B|$. At fixed nonzero gas <pressure>, stronger field increases <surface coverage>; in an isotherm plot, the half-coverage <pressure> $p_{1/2}=1/b$ moves to lower <pressure>. The saturation coverage remains one because the number of sites is unchanged.
For weak fields, with $\theta_0=b(0)p/[1+b(0)p]$,
$$
\theta(B)=\theta_0+\frac12\theta_0(1-\theta_0)x^2+O(x^4).
$$
There is no linear small-field change. At large $|x|$, $\cosh x\sim e^{|x|}/2$, and at fixed $p>0$ the coverage approaches one. The field lowers the <free energy> of the surface <spin> ensemble relative to the unsplit bulk; simply averaging the two surface energies to $-E_b$ would miss this equilibrium effect.
\Image[/past-exam-of-the-chemistry-a-course-of-the-university-of-cambridge/2022/paper-2-magnetic-adsorption.png]
{title=Original magnetic Langmuir isotherms and field-dependent coverage in dimensionless variables}
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