Model of cardinality $\lambda$ where every definable subset is either finite or has size $\lambda$ Let $T$ be consistent and $P$ is a predicate in $L(T)$. Suppose $\lambda \ge \aleph_0 + |L(T)|$. Then if $T$ has an infinite model, there is a model $M$ of $T$ of cardinality $\lambda$ such that every definable subset is finite or has cardinality $\lambda$.
How does one prove this? First of all, suppose $T$ does not prove $P$ is finite (in which case then there's no much to be done wtih $P$). Then $T \cup \{P(c_i) : i < \lambda \} \cup \{\neg P(d_i) : i < \lambda \}$ is consistent where $\{c_i, d_i : i < \lambda\}$ are new constant symbols. This leaves the problem of, well, all the other formulas. There may be other relation, constant and function symbols in $L(T)$, which I'm not sure how to deal with. Is it a matter of taking every non algebraic formula and saying there are $\lambda$ many constants that satisfy it, and $\lambda$ many that don't? If so, then what does $P$ have to do with it at all?
Edit: I forgot to make sure that the model would turn out with cardinality $\lambda$. But this simply follows from the fact that since we'll be adding $\lambda$ many new constant symbols for every formula $\varphi$ and there are $\leq \lambda$ many such formulas, the total number of constant symbols added will be $\lambda \times \lambda = \lambda$. So we may find a model of cardinality $\lambda$ by downward lowenheim skolem once we apply compactness.
 A: $P$ is irrelevant in the statement, so let's forget about it.
What you wrote is almost right. If $\varphi(x)$ is a non-algebraic formula with one free variable, then it's consistent to add $\lambda$-many new constant symbols and say that they all satisfy $\varphi(x)$. But it's not necessarily consistent to add $\lambda$-many new constant symbols and say they all don't satisfy $\varphi(x)$, unless $\lnot \varphi(x)$ is also non-algebraic! So it's better to just handle $\varphi$ and $\lnot \varphi$ separately.
Now the details of the construction depend on whether "definable set" means definable with parameters or not. Let's assume we want to handle definable sets with parameters, since this is a little bit trickier.
Start with a model $M_0\models T$ of cardinality $\lambda$ (which exists by Löwenheim-Skolem). Let $T_M$ be the elementary diagram of $M$ (the complete theory of $M$ in the language $L_M$ with a constant symbol for every element of $M$ - note that this language has cardinality $\lambda$).
Make a list of all the non-algebraic formulas in one variable with parameters from $M$. There are $\lambda$-many of these. For each one, say $\varphi(x,\overline{a})$, introduce $\lambda$-many new constant symbols and add to $T_M$ the axiom $\varphi(c,\overline{a})$ for each new constant $c$, as well as the axioms $c\neq d$ for each pair of new constants $c$ and $d$. The resulting language still has cardinality $\lambda$, so by Löwenheim-Skolem, the resulting theory has a model $M_1$ of cardinality $\lambda$, and $M_0\preceq M_1$ since $M_1\models T_M$.
Now $M_1$ is a model of $T$ of cardinality $\lambda$ such that every set definable  with parameters from $M_0$ is finite or has cardinality $\lambda$. But what about sets definable with parameters from $M_1$ which are not in $M_0$? To deal with these, we repeat the construction above, building an elementary chain $M_0\preceq M_1\preceq M_2\preceq \dots$. The union of this chain is a model of cardinality $\lambda$ such that every definable set with parameters is finite or has cardinality $\lambda$. Why? The finitely many parameters all appear in some $M_n$, so already the definable set has size $\lambda$ in $M_{n+1}$, and it can only grow when we take the union.
If you only care about definable sets without parameters, then there's no need for the elementary chain: you're already done after the first step.
