= Solution
Order the collection $\operatorname{Ch}(P)$ of all <chains in a partial order>, including the empty chain, by inclusion. Suppose that $f:\operatorname{Ch}(P)\to P$ were an <order-preserving> injection. The <chain-poset nonembedding lemma> follows by a <transfinite recursion> of length the <Hartogs ordinal> $h(P)$:
$$
p_\alpha=f(C_\alpha),\qquad C_\alpha=\{p_\beta:\beta<\alpha\}.
$$
Inductively, the previously constructed points form a strictly increasing chain. If $\beta<\alpha$, then $C_\beta\subsetneq C_\alpha$, since $p_\beta$ belongs to the latter but not the former. Monotonicity gives $p_\beta\leq p_\alpha$, and injectivity makes this inequality strict. Thus the recursion continues at every $\alpha<h(P)$ and gives an injection $h(P)\to P$, contradicting the defining property of the <Hartogs ordinal>. Therefore $\boxed{\operatorname{Ch}(P)\text{ has no order-preserving embedding into }P.}$ The empty chain is part of the convention: excluding it would make a one-point poset a counterexample.
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