= Solution
In the observer's <rest frame>, $U=(c,0)$ and $P=(E'/c,\mathbf p')$. The <Lorentz invariance> of the <Minkowski inner product> gives $P\cdot U=E'$ and
$$
P\cdot P=\frac{E'^2}{c^2}-|\mathbf p'|^2=m^2c^2.
$$
The relation between <four-momentum> and <velocity> is $\mathbf p'=E'\mathbf v/c^2$, so
$$
\frac{v^2}{c^2}=\frac{c^2|\mathbf p'|^2}{E'^2}=1-\frac{c^2(P\cdot P)}{(P\cdot U)^2}.
$$
Taking the nonnegative square root proves
$$
\boxed{v=c\sqrt{1-\frac{c^2(P\cdot P)}{(P\cdot U)^2}}.}
$$
For a future-directed massive particle and observer, $P\cdot U>0$ and the expression lies in $[0,1)$; the massless limit, with nonzero energy, gives $v=c$.
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