= Solution
\b[True.] Each element of the <Galois group> preserves the extending <absolute value on a field>, because composing it with that <absolute value on a field> gives another extension, equal to the original by part (a). Thus all conjugates of $x$ have the same <absolute value on a field>. Writing $d=[K:\mathbb Q_p]$, the <field norm> gives
$$
|N_{K/\mathbb Q_p}(x)|_p
=\prod_{\sigma\in\operatorname{Gal}(K/\mathbb Q_p)}|\sigma x|
=|x|^d.
$$
A <unit> <field norm> has <absolute value on a field> one, forcing $|x|=1$. This means both $x$ and $x^{-1}$ are in the <valuation ring>, so $x$ is a <unit>.
Equivalently, with normalized <discrete valuations> and <residue degree> $f$,
$$
\boxed{v_p(N_{K/\mathbb Q_p}x)=f\,v_K(x)=0
\quad\Longrightarrow\quad v_K(x)=0.}
$$
This is <unit norm detects units in a local extension>.
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