= Solution
Use $\hbar=k_{\rm B}=1$ and dimensionless spin operators. At one site, the <canonical anticommutation relations> imply the <on-site spin-square normal-ordering identity>
$$
\mathbf S_n^2=\frac34(n_n-2n_{n\uparrow}n_{n\downarrow}),\qquad
:\mathbf S_n^2:=\mathbf S_n^2-\frac34n_n=-\frac32n_{n\uparrow}n_{n\downarrow}.
$$
Indeed the empty and doubly occupied states have spin zero, while either singly occupied state has $\mathbf S^2=3/4$. Thus the interaction in normal-ordered form is \b[$3U\sum_n c^\dagger_{n\uparrow}c^\dagger_{n\downarrow}c_{n\downarrow}c_{n\uparrow}$]. There is no residual $3Un/2$ chemical-potential shift: that contraction has cancelled.
Insert <fermionic coherent state> resolutions between imaginary-time evolution factors. With $\xi_k=\epsilon_k-\mu$ and $D_n=\bar\psi_{n\uparrow}\bar\psi_{n\downarrow}\psi_{n\downarrow}\psi_{n\uparrow}$, the <grand-canonical coherent-state action> is
$$
\boxed{Z=\int_{\rm AP}\mathcal D(\bar\psi,\psi)\,e^{-S_0-3U\int_0^\beta d\tau\sum_nD_n},\qquad
S_0=\int_0^\beta d\tau\sum_{k\sigma}\bar\psi_{k\sigma}(\partial_\tau+\xi_k)\psi_{k\sigma}.}
$$
The independent barred and unbarred <Grassmann variables> satisfy antiperiodic boundary conditions at $0,\beta$. Adjacent time-slice labels in the normal-ordered symbol fix the continuum prescription and thermal normalization. The interaction coefficient is $3U$, not the $U$ of a differently normalized <Hubbard model>.
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