= Solution
Two unary predicates suffice. Require all four <unary predicate types> to be inhabited:
$$
\boxed{
\exists x\,(P(x)\land Q(x))\ \land\
\exists x\,(P(x)\land\neg Q(x))\ \land\
\exists x\,(\neg P(x)\land Q(x))\ \land\
\exists x\,(\neg P(x)\land\neg Q(x)).}
$$
There are no constants, no <logical equality> and no nonunary predicates. Any two of these conditions disagree on at least one predicate, so one domain element cannot witness both. Hence every satisfying <first-order structure> has at least four elements, by <distinct witnesses from incompatible unary predicates>. Conversely take domain $\{a,b,c,d\}$, $P=\{a,b\}$ and $Q=\{a,c\}$: the four elements realize the four displayed <unary predicate types>, respectively. These element names describe the example; they are not constants in the <first-order sentence>.
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