= Solution
Order the collection $\operatorname{Ch}(P)$ of every <chain in a partial order>, including the empty chain, by inclusion. Suppose that $j:\operatorname{Ch}(P)\to P$ is an <order-preserving function> and an <injection>. Let $\theta=h(P)$ be the <Hartogs ordinal>: by <Hartogs theorem>, no <injection> from $\theta$ to $P$ exists.
Use <transfinite recursion> to define, for $\alpha<\theta$,
$$
C_\alpha=\{p_\beta:\beta<\alpha\},\qquad p_\alpha=j(C_\alpha).
$$
We verify both that $j(C_\alpha)$ is defined and that the values are strictly increasing. Assume inductively that the earlier values are strictly increasing. Then $C_\alpha$ is a <chain in a partial order>. For any $\beta<\alpha$, $C_\beta\subseteq C_\alpha$, so the fact that $j$ is <order-preserving> gives $p_\beta\le p_\alpha$. Furthermore $p_\beta\in C_\alpha$, whereas $p_\beta\notin C_\beta$ by distinctness of the earlier values. Thus $C_\beta\ne C_\alpha$, and <injectivity> gives $p_\beta\ne p_\alpha$. We have proved
$$
\beta<\alpha\quad\Longrightarrow\quad p_\beta<p_\alpha.
$$
The empty initial chain starts the induction, and the same argument applies at every <limit ordinal>.
If a globally defined recursion rule is desired, apply $j$ when the earlier range is a chain and otherwise use the fixed value $j(\varnothing)$. The induction shows that the fallback case never occurs. Thus no choices of new points are being made: the <function> $j$ uniquely determines every stage.
The resulting <function> $\alpha\mapsto p_\alpha$ is an <injection> from $\theta$ to $P$, contrary to its defining <Hartogs ordinal> property. Hence the <chain-poset nonembedding lemma> gives
$$
\boxed{\text{No order-preserving injection }\operatorname{Ch}(P)\longrightarrow P\text{ exists}.}
$$
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