= Solution
Write $M=\mathcal N=V/b$ for the number of cells; this is fixed even though the <particle> number $N$ fluctuates. Approximate each position <integral> by the cell sum $\int d^3r_i\simeq b\sum_{\alpha_i=1}^M$, evaluating the slowly varying intercell <pair potential> at cell centres. With the assumed single-occupancy restriction, the fixed-$N$ contribution becomes
$$
\frac1{N!}\left(\frac{e^{\beta\mu}b}{\lambda^3}\right)^N
\sum_{\alpha_1,\ldots,\alpha_N\ \mathrm{distinct}}
\exp\left[-\frac\beta2\sum_{i\ne j}\phi_{\alpha_i\alpha_j}\right].
$$
An occupation configuration $\{n_\alpha\}$ with $N=\sum_\alpha n_\alpha$ selects an unordered set of $N$ occupied cells. There are exactly $N!$ ways to assign the temporary <particle> labels to that set. They all have the same interaction <energy>, cancelling the prefactor $1/N!$.
Summing over all <particle> numbers is therefore equivalent to summing each cell occupation independently over $0,1$:
$$
\boxed{\Xi\simeq\sum_{n_1=0}^1\cdots\sum_{n_M=0}^1
\left(\frac{e^{\beta\mu}b}{\lambda^3}\right)^{\sum_\alpha n_\alpha}
\exp\left[-\frac\beta2\sum_{\alpha\ne\alpha'}n_\alpha n_{\alpha'}\phi_{\alpha\alpha'}\right].}
$$
The activity per cell, $e^{\beta\mu}b/\lambda^3$, is dimensionless. This is a <lattice gas>, with each cell either empty or occupied.
The exclusion of multiple occupation is an additional physical modeling assumption, explicitly imposed by the question. Making a cell small does not by itself forbid two point <particles> from entering it; a repulsive core or a dilute-occupancy approximation would be needed to justify that restriction starting from a general continuum <pair potential>. In the remaining parts it is part of the defined cell model.
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