= Solution
The quadratic part of the covariant-derivative term contains $g^2v^2A_\mu A^\mu/2-gvA_\mu\partial^\mu b$. Its mixing term cancels the $+gvA_\mu\partial^\mu b$ in the stated <gauge fixing> term. Thus the vector <mass> and diagonal quadratic action are
$$
\boxed{M_A^2=g^2v^2,\qquad M_A=|gv|,}
\qquad
\mathcal L_{A,2}=-\frac14F^2+\frac12M_A^2A^2-\frac1{2\xi}(\partial\cdot A)^2.
$$
For nonzero $p^2$ introduce the transverse and longitudinal <transverse and longitudinal momentum projectors> $P_T^{\mu\nu}=g^{\mu\nu}-p^\mu p^\nu/p^2$ and $P_L^{\mu\nu}=p^\mu p^\nu/p^2$. After integration by parts, the inverse quadratic kernel is
$$
K^{\mu\nu}=-(p^2-M_A^2)P_T^{\mu\nu}
-\frac{p^2-\xi M_A^2}{\xi}P_L^{\mu\nu}.
$$
Their orthogonality makes the inversion immediate. The <gauge-boson propagator> is
$$
\boxed{D_A^{\mu\nu}(p)=-i\left[
\frac{P_T^{\mu\nu}}{p^2-M_A^2+i0}
+\frac{\xi P_L^{\mu\nu}}{p^2-\xi M_A^2+i0}\right].}
$$
Equivalently,
$$
D_A^{\mu\nu}(p)=\frac{-i}{p^2-M_A^2+i0}
\left[g^{\mu\nu}-(1-\xi)\frac{p^\mu p^\nu}{p^2-\xi M_A^2+i0}\right].
$$
The projector formula explains the physical massive pole and the gauge-dependent longitudinal pole. The apparent $1/p^2$ in the projectors cancels in the equivalent expression; it is not an extra massless physical state. At $\xi=1$ the numerator reduces to $g^{\mu\nu}$. At fixed <momentum> the $\xi\to\infty$ limit gives the <unitary gauge> <propagator> $-i(g^{\mu\nu}-p^\mu p^\nu/M_A^2)/(p^2-M_A^2+i0)$. The relation between this limit and ultraviolet power counting is not uniform in <momentum>.<Gauge Theory by David Tong>[§2.7.3, Higgs-phase gauge-boson masses and the distinction between gauge descriptions and physical spectrum]{ref}
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