= Solution
Let $v=\sigma^2>0$, $d_k=k+1$, and $S_k(a)=\|y-X_ka\|^2$. Under the standard interpretation that the two stated priors are independent, multiply the normal likelihood by the normal coefficient prior and the <inverse-gamma distribution> density. The <independent normal and inverse-gamma regression priors> give
$$
\boxed{\pi(a_k,v\mid x,y)\propto
v^{-(\alpha+1+n/2)}
\exp\left[-\frac{\beta+S_k(a_k)/2}{v}
-\frac12(a_k-\mu_k)^T\Sigma_k^{-1}(a_k-\mu_k)\right],\quad v>0.}
$$
For a fixed model the determinant and Gaussian normalizing constants are independent of $a_k,v$ and may be omitted here; $\Sigma_k$ is assumed <positive-definite>. The coefficient prior is not scaled by $v$, so replacing its precision by $\Sigma_k^{-1}/v$ would describe a different prior.
For an additional check, completing the coefficient square and reading the <variance> kernel give the full conditional distributions
$$
a_k\mid v,x,y\sim N(m_k,V_k),\quad
V_k=(X_k^TX_k/v+\Sigma_k^{-1})^{-1},\quad
m_k=V_k(X_k^Ty/v+\Sigma_k^{-1}\mu_k),
$$
and
$$
v\mid a_k,x,y\sim\operatorname{InvGamma}(\alpha+n/2,\beta+S_k(a_k)/2).
$$
Marginal priors alone would not determine their joint prior without the independence assumption. The front-page inverse-gamma mean has a typographical error: this density has mean $\beta/(\alpha-1)$ when $\alpha>1$, not a denominator $\alpha+1$. The density itself, used in the posterior calculation, fixes the correct parameterization.
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