 Solution

Order the collection $\mathrm{Ch}(P)$ of all chains in a partial order, including the empty chain, by inclusion. Suppose that $f:\mathrm{Ch}(P)\to P$ were an order-preserving injection. The chain-poset nonembedding lemma follows by a transfinite recursion of length the Hartogs ordinal $h(P)$:Inductively, the previously constructed points form a strictly increasing chain. If $\beta <\alpha$, then $C_{\beta }⊊C_{\alpha }$, since $p_{\beta }$ belongs to the latter but not the former. Monotonicity gives $p_{\beta }\le p_{\alpha }$, 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 $\boxed{\mathrm{Ch}(P)\text{ has no order-preserving embedding into }P.}$ The empty chain is part of the convention: excluding it would make a one-point poset a counterexample.