Jech's Axiom of Choice book, exercise 4.12 is that the Selection Principle (for all $\mathcal{F}$ there is a function $f: \mathcal{F} \rightarrow V$ such that $|X| \ge 2 \rightarrow \emptyset \subsetneq f(X) \subsetneq X$) is equivalent to K(1) (every set can be injected into $\mathcal{P}(Ord)$). He remarks, "Consequently, the Selection Principle implies the Ordering Principle [every partial order can be extended to a total order]." This isn't obvious to me.
My guess is the intended proof is something like this: "Let $(P, \le)$ be a partial order. We may assume $P=\mathcal{P}(\alpha),$ for some ordinal $\alpha.$ There is some canonical way to strictly extend the ordering. By transfinite iteration, there is a canonical total extension of the ordering." I can't tell what this canonical way to extend $(P,\le)$ is. My first guess was to find the minimal $\beta<\alpha$ such that two incomparable elements differ at $\beta,$ and for all pairs $p \perp q,$ where $\beta \in q \setminus p,$ declare $p<q.$ Unfortunately, this can create a cycle in the order. I'm not sure what else to try.
You are misreading things.
The ordering principle states "Every set can be linearly ordered". The order extension principle states "every partial order can be extended to a linear order".
If every set can be injectively mapped into $\mathcal P(\alpha)$ for some ordinal $\alpha$, then it will inherit the linear ordering definable on $\mathcal P(\alpha)$.
Interestingly enough, I cannot find any work separating Kinna–Wagner's Selection Principle (Or $\sf KW$ or $\sf KWP_1$ as it is sometimes denoted) from the Order Extension Principle. Either the non-implication is known, or the question is still open, as indicated by the diagrams in Herrlich's "Axiom of Choice" book.