= Solution
At a stationary symmetry-breaking <vacuum> of a gauge-invariant scalar potential, the angular field has zero potential curvature. Its free <kinetic term> is $(\partial b)^2/2$, while the displayed gauge-fixing term adds $-\xi g^2v^2b^2/2$. Hence
$$
\boxed{D_b(p)=\frac{i}{p^2-\xi M_A^2+i0},\qquad m_b^2=\xi M_A^2.}
$$
This is the would-be <Goldstone boson>, whose gauge-dependent pole accompanies the longitudinal pole in <3/b/i>; it is not an additional physical scalar particle of that <mass>. For $M_A\ne0$, fixed finite $p$ and $\xi\to+\infty$,
$$
D_b(p)\sim-\frac{i}{\xi M_A^2}\longrightarrow0.
$$
The angular field becomes infinitely heavy in this gauge and is absent from the <unitary gauge> propagating description. The qualification about fixed <momentum> matters: loop integrations probe arbitrarily large $p$, so one cannot automatically interchange the $\xi\to\infty$ limit with ultraviolet integration. If $gv=0$, there is no massive Higgs phase and the asserted heavy-field limit does not follow. A separate radial <Higgs boson> may remain; its <mass> depends on the unspecified potential and is not determined by this gauge-fixing term.
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