= Solution
Use the shape-scale convention for the <inverse-gamma distribution>: $\operatorname{IG}(a,b)$ has density proportional to $s^{-a-1}e^{-b/s}$ for $s>0$. Write $X=X_k$, $\beta=\beta_k$, $\mu=\mu_k$, $\Sigma=\Sigma_k$ and $s=\sigma^2$. The <independent normal and inverse-gamma regression priors> make the conditional <log-likelihood> plus log prior in $\beta$ equal, up to a constant, to
$$
-\frac12\left[\beta^T(X^TX/s+\Sigma^{-1})\beta
-2\beta^T(X^Ty/s+\Sigma^{-1}\mu)\right].
$$
Define $V=(X^TX/s+\Sigma^{-1})^{-1}$ and $m=V(X^Ty/s+\Sigma^{-1}\mu)$. <Completing the square> turns this expression into $-(\beta-m)^TV^{-1}(\beta-m)/2$ plus a constant. Thus the conditional <posterior distribution> is the <multivariate normal distribution>
$$
\boxed{\beta_k\mid\sigma^2,x,y\sim N_{k+1}(m,V),\quad
V=(X_k^TX_k/\sigma^2+\Sigma_k^{-1})^{-1},\quad
m=V(X_k^Ty/\sigma^2+\Sigma_k^{-1}\mu_k).}
$$
Positive definiteness of the prior <covariance matrix> makes this well defined even if the <design matrix> is rank deficient.
Holding $\beta$ fixed, the factors depending on $s$ are
$$
s^{-(a+n/2)-1}\exp\left(-\frac{b+\|y-X\beta\|^2/2}{s}\right).
$$
Consequently
$$
\boxed{\sigma^2\mid\beta_k,x,y\sim\operatorname{IG}\left(a+n/2,\ b+\tfrac12\|y-X_k\beta_k\|^2\right).}
$$
There is no additional $(k+1)/2$ in the shape: the <normal distribution> prior for $\beta_k$ is independent of $\sigma^2$. A prior scaled by $\sigma^2$ would be a different model and would change that calculation.
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