Let $\mathcal{K}$ be a class of finite $\mathcal{L}$-structures and $T=\text{Th}(\mathcal{K})$.
If $T$ admits an infinite model then $T$ admits finite models of arbitrarily large cardinal.
Let suppose that $T$ has an infinite model. But as $T=\text{Th}(\mathcal{K})$ we can deduce that $T$ also admits finite models. Then I do not know how to make a link between the infinite model and the finite models of $T$ ?
Notice that, at the end we want to prove that for all $n\in \mathbb{N}$ there exists a model of $T$, $\mathfrak{R}$ such that $\vert R \vert\ge n$.
Thanks in advance !
Every element of the class $\mathcal{K}$ is a finite model of the theory $T=\mathrm{Th}(\mathcal{K})$. So if $\mathcal{K}$ contains finite structures of arbitrarily large cardinality, then $T$ admits finite models of arbitrarily large cardinality, and we're done.
Otherwise, there is a bound $B \in \mathbb{N}$ so that the class $\mathcal{K}$ does not contain finite structures of size larger than $B$. But then take a first-order sentence $\varphi$ which states "the structure has no more than $B$ elements". Clearly, this sentence $\varphi$ holds in every structure in the class $\mathcal{K}$, and consequently $\varphi$ belongs to the theory $T$. But an infinite model of $T$ would have more than $B$ elements, which would contradict $\varphi$. Therefore, $T$ does not have infinite models.
Taking contrapositives, we get that if $T$ does have an infinite model, then $T$ also has finite models of arbitrarily large cardinality (and the class $\mathcal{K}$ that we started with contains such models).