= Solution
Let $\Lambda=h/\sqrt{2\pi mk_{\rm B}T}$ be the <thermal de Broglie wavelength>. The two free momentum components of one adsorbed atom give the phase-space integral
$$
q_{\rm trans}^{(2)}=\frac1{h^2}\int_{\sigma}d^2r\int_{\mathbb R^2}dp_x\,dp_y\,
e^{-\beta(p_x^2+p_y^2)/(2m)}
=\frac{\sigma}{h^2}(2\pi mk_{\rm B}T).
$$
Each momentum integral is a <Gaussian integral>, so
$$
\boxed{q_{\rm trans}^{(2)}=\frac{\sigma}{\Lambda^2},\qquad
Z_N^{(2)}=\frac1{N!}\left(\frac{\sigma}{\Lambda^2}\right)^N.}
$$
The second expression is the <classical ideal-gas partition function> for mobile, indistinguishable surface atoms. Its validity requires a dilute nondegenerate gas, $N\Lambda^2/\sigma\ll1$, and neglects interactions and excited motion normal to the surface. The $N!$ appears here because particles move between positions, unlike the fully occupied fixed-site model of A6.
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