So I was going through this paper and under Uncertainty modeling it says 
So I tried deriving it on my own and I got
$p(\omega | X, Y) = \frac{p(Y | X, \omega) \cdot p(X,\omega)}{p(Y | X) \cdot P(X)}$
I am not quite sure how they removed the term $p(X,\omega)$ in the numerator and $P(X)$ term in the denominator.
Another doubt is regarding how they arrived at the integral for
$p(y*|x*,X,Y) = \int p(y*|x*,\omega) \cdot p(\omega|X,Y) d\omega$
is there some proof for it or is it based on intuition?
I would say that the logic is the following:
$$ p(\omega \lvert X,Y) = \frac{p(Y\lvert X, \omega)p(X, \omega)}{\int d\omega \, p(Y\lvert X, \omega)p(X, \omega)} $$ Now because the data $X$ and the parameters $\omega$ are independent from each other, you can use $p(X, \omega) = p(X)p(\omega)$. And I assume somehow you do not integrate over $X$ because these are fixed variables more than random variables (I believe), furthermore this would also imply $p(X) = 1$.
Therefore we are left with:
$$ p(\omega \lvert X,Y) = \frac{p(Y\lvert X, \omega)p(\omega)}{\int d\omega \, p(Y\lvert X, \omega)p(\omega)} \equiv \frac{p(Y\lvert X, \omega)p(\omega)}{ p(Y\lvert X)} $$
Hope someone more knowledgable can help :)
