 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 $\int d^3{r}_i\simeq b{\sum }_{{\alpha }_i=1}^M$, evaluating the slowly varying intercell pair potential at cell centres. With the assumed single-occupancy restriction, the fixed-$N$ contribution becomesAn 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 $0,1$: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.