So there are a few questions that have asked this before here and here, but I seem to be missing a step.
$$ \begin{aligned} p(f_*|x_*,X,y) &= \int p(f_*,w|x_*,X,y)~dw \quad \text{(marginalise $w$)}\\ &= \int p(f_*|x_*,X,y,w)p(w|x_*,X,y)~dw \quad \text{(chain rule)}\\ &= \int p(f_*|x_*,w)p(w|X,y)~dw \quad \text{($f_* \mathrel{\unicode{x2AEB}} X, y$ given $w$ and $w \mathrel{\unicode{x2AEB}} x_*$)}\\ \end{aligned} $$
Now, I don't understand what distribution $p(f_*|x_*,w)$ is, isn't it just a constant when both {$x_*, w$} are given? There seems to be some step I'm missing after substituting $f_* = x_*^Tw$ and then solving the integral:
$$ \begin{aligned} p(f_*|x_*,X,y) &= \int p(f_*|x_*,w)p(w|X,y)~dw\\ &= \int p(x_*^Tw|x_*,w)p(w|X,y)~dw \quad \text{($f_* = x_*^Tw$)}\\ &= \textbf{what goes here?}\\ &= x_*^T\mathcal{N}\left(\frac{1}{\sigma_n^2} A^{-1}Xy, A^{-1}\right) \quad \text{(is this right?)}\\ &= \mathcal{N}\left(\frac{1}{\sigma_n^2} x^T_*A^{-1}Xy, x_*^TA^{-1}x_*\right) \end{aligned} $$