= Solution
Assume that $\kappa$ is regular and uncountable in $M$, as required for this claim. If an antichain $A\in M$ had size $\kappa$, apply the <delta-system lemma at a regular uncountable cardinal> inside $M$ to the finite domains of its conditions. After thinning, their domains form a delta-system with finite root $R$.
There are fewer than $\kappa$ possible restrictions to $R$, because each coordinate $(\alpha,n)\in R$ has fewer than $\kappa$ possible values. Since $\kappa$ is a <regular cardinal>, we can thin again to $\kappa$ conditions agreeing on $R$. Any two now have disjoint domains outside $R$ and agree on $R$, so their union is a common extension. This contradicts that $A$ is an antichain. Therefore
$$
\boxed{\mathbb P\text{ has the }\kappa\text{-chain condition}.}
$$
If the word “regular” is allowed to include $\kappa=\omega$, the printed claim needs the additional hypothesis that $\kappa$ is uncountable; the finite-condition order can have an infinite antichain when $\kappa=\omega$.
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