Let $\alpha: I \mapsto \mathbb{R^3}$ be a helix and $u$ be the constant unit vector whose angle $\theta$ with $\alpha'(t)$ is constant. Let $s(t)$ be the arclength function of $\alpha$ starting at $t = 0$. Consider the curve $\beta(t) = \alpha(t) - s(t)\cos(\theta)u$ and prove that:
a) $\beta$ is contained in the plane that goes through $\alpha(0)$ and is orthogonal to $u$
b) Prove that $K_\beta = \cfrac{K_\alpha}{\sin^2(\theta)}$
b) is easy, but I'm having some trouble with a). Everything I tried got hairy pretty fast. I think there might be a typo and I should consider $\beta'$ as the curve contained in the mentioned plane, 'cause I've already proven it in that case and it makes much more sense, but I'm not sure.
Progress: I think I got it. Assuming, WLOG, that $\alpha$ is a unit speed curve, since $\beta'(t) \cdot u =( \alpha'(t) - s'(t)\cos(\theta)u)\cdot u = \cos(\theta)-\cos(\theta) = 0$, we can integrate and find that $\beta(t) \cdot u = c$... and for anyone seeing this who wants the answer, see the discussion in the comments, because the previous update had an error.
Your computation isn't quite right, as $\alpha'(t)\cdot u = s'(t)\cos\theta$. (Recall that $s'(t)$ is the speed with which the particle moves.) [You also never defined $\theta$ in your question!]
Note that $s(t)$ is the arclength starting at $t=0$, so $s(0)=0$. This shows that $\beta(0)=\alpha(0)$. Now, can you show that $\big(\beta(t)-\alpha(0)\big)\cdot u = 0$ for all $t$?