 Solution

Take a countable elementary substructure ${\mathcal{B}}^{\prime }$ of ${\mathcal{A}}^{\prime }$ containing the named constant, using the Downward Lowenheim-Skolem theorem. Its reduct $\mathcal{B}$ has infinitely many singleton and doubleton classes. In particular it contains the partner of each of its non-singleton elements, and its interpretation of $P$ is exactly the singleton-class predicate.

Choose new distinct elements $u,v\notin B$. DefineIn $\mathcal{C}$, make $u$ a new singleton class. In $\mathcal{D}$, make $\{u,v\}$ a new doubleton class; keep every old class of $\mathcal{B}$ unchanged. All three reducts are first-order models of the complete theory, and $\mathcal{B}\subseteq \mathcal{C}\subseteq \mathcal{D}$.

Expand $\mathcal{C},\mathcal{D}$ by interpreting $P$ as singleton-class membership and retaining the original named constant from $B$. The resulting countable first-order structures ${\mathcal{C}}^{\prime },{\mathcal{D}}^{\prime }$ are each isomorphic to ${\mathcal{B}}^{\prime }$: all have infinitely many classes of each size, and the bijections of classes can be chosen to preserve the constant and its class. Hence they are first-order models of $\mathrm{Th}({\mathcal{A}}^{\prime })$. Both inclusions ${\mathcal{B}}^{\prime }\subseteq {\mathcal{C}}^{\prime }$ and ${\mathcal{B}}^{\prime }\subseteq {\mathcal{D}}^{\prime }$ are genuine structure embeddings, since no class meeting $B$ was changed. The given quantifier elimination, together with the preceding implication to first-order model completeness, yieldsTaking reducts gives $\boxed{\mathcal{B}⪯\mathcal{C},\mathcal{B}⪯\mathcal{D}}$. Butso $\boxed{\mathcal{C}⋠\mathcal{D}}$. There is no conflict with quantifier elimination in the expanded language: $P(u)$ changes truth value, so ${\mathcal{C}}^{\prime }$ is not a substructure of ${\mathcal{D}}^{\prime }$.