Geometry of buta-1,3-dien-1-ol
Let’s start off with the things you thought you got right: Oxygen will not be $\mathrm{sp^3}$ hybridised but somewhat closer to $\mathrm{sp^2}$. This is because that is how the oxygen lone pair in a p-type orbital can take part in &#x3c0; conjugation. I’m not sure what $\angle(\ce{C-O-H})$ I would expect but it’ll be somewhere between $109^\circ$ and $120^\circ$, that’s for sure.

To the entire molecule’s structure: Yes, at first glance it seems like a potato. However, there is still something that you can say: for proper conjugation of the double bonds (usually favourable), you need a planar structure. Thus, the entire molecule will be planar with all atoms in the same plane. That is about all that you can safely derive, since there can still be rotation around the $\ce{C-C}$ single bond.

Also, as Ivan pointed out, the molecule is likely to prefer the keto-tautomer rather than the enol one. The conjugation argument may apply, but it is rather weak.