No, you cannot marginalize the prior and then multiply the marginal with g(x). Here is why:

The correct way to use Bayes theorem is to do the following (also suggested by John):

$g_{p1}(x,y,z) \propto g(x,y,z) k(x,y,z)$

Thus,

$g_{p1}(x) \propto \int_{y,z} \bigl(g(x,y,z) k(x,y,z) \bigr)$

Your want to do the following:

$g_{p2}(x) \propto \int_{y,z} \bigl(g(x,y,z) \bigr) \int_{y,z} \bigl(k(x,y,z) \bigr)$

In general, $g_{p1}(x)$ and $g_{p2}(x)$ will not be identical.