For homeomorphic graphs one can disregard vertices of degree 2
(forget the vertex was ever there, but keep the edges). If you do that here, you get a graph with 4 vertices only: a,b,g,h. So this cannot be homeomorphic to K(3,3).
We clearly see that it cannot be homeomorphic to K(3,3) as we need atleast 3 edges from each verted connecting to the other set of vertices.
