= Solution
Write $v=\sigma^2$, $d=k+1$, $X=X_k$, $a=a_k$, $\mu=\mu_k$ and $\Sigma=\Sigma_k$, with $\Sigma$ <positive-definite>. For fixed $v$, the terms involving $a$ in the posterior exponent are
$$
-\frac12\left[a^T\left(\frac{X^TX}{v}+\Sigma^{-1}\right)a-2a^T\left(\frac{X^Ty}{v}+\Sigma^{-1}\mu\right)\right].
$$
Complete the square. With $V_k=(X_k^TX_k/v+\Sigma_k^{-1})^{-1}$ and $m_k=V_k(X_k^Ty/v+\Sigma_k^{-1}\mu_k)$, this gives
$$
\boxed{a_k\mid v,x,y\sim N_{k+1}(m_k,V_k).}
$$
For fixed $a_k$, collect the powers and exponential terms involving $v$:
$$
\pi(v\mid a_k,x,y)\propto v^{-(\alpha+n/2)-1}\exp\left[-\frac{\beta+\tfrac12\|y-X_ka_k\|^2}{v}\right].
$$
Hence, in the paper's shape-scale <inverse-gamma distribution> convention,
$$
\boxed{v\mid a_k,x,y\sim\operatorname{IG}\left(\alpha+\frac n2,\beta+\frac12\|y-X_ka_k\|^2\right).}
$$
There is no extra $(k+1)/2$ in this shape, because the normal coefficient prior is independent of $v$. These are the <independent normal and inverse-gamma regression priors> conditionals.
The Gaussian normalization printed in the PDF is incorrect. The proper $d$-dimensional prior <probability density function> is $(2\pi)^{-d/2}|\Sigma_k|^{-1/2}\exp[-(a_k-\mu_k)^T\Sigma_k^{-1}(a_k-\mu_k)/2]$. For fixed $k$ this correction is constant in $a_k,v$ and leaves the two conditionals unchanged, but it must be retained when comparing different model orders in part (b).
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