= Solution
For $P^\mu=(E/c,\mathbf p)$, directly contract the supplied vector:
$$
W\cdot P=(\mathbf s\cdot\mathbf p)\frac Ec
-\frac Ec\mathbf s\cdot\mathbf p=0.
$$
Thus $\boxed{U_W\cdot P=0}$ whenever its stated normalization is defined, namely $m_0>0$ and $|\mathbf s|>0$. This is the <Pauli-Lubanski orthogonality> property.
The description as a unit vector, however, needs an additional assumption. From the given components and the mass shell,
$$
W^2=-m_0^2c^2|\mathbf s|^2-|\mathbf s\times\mathbf p|^2,\qquad
\boxed{U_W^2=-1-\frac{|\mathbf s\times\mathbf p|^2}{m_0^2c^2|\mathbf s|^2}.}
$$
It equals minus one only for <spin> parallel to <momentum>, including the zero-momentum case. A concrete counterexample is $\mathbf s=(s,0,0)$ and $\mathbf p=(0,m_0c,0)$: the proposed normalized vector has square minus two. Orthogonality is still true. A genuine <Pauli-Lubanski pseudovector> constructed by boosting a fixed rest-frame <spin> has invariant norm $-m_0^2c^2|\mathbf s_{\mathrm{rest}}|^2$; the source's generic spatial expression and denominator do not supply that normalization for arbitrary $\mathbf s$.
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