This question has an answer here: Curvature of parallel curve for a regular plane curve but I do not understand it for two reasons, 1) being that I don't see the justification for $\frac{ds}{dt} = 1/k$ and also 2) by taking absolute values at the end, he actually loses the sign needed for the result.
With that out of the way, the question is to prove that for a parallel curve $\alpha_\lambda(t) = \alpha(t) + \lambda n(t)$ that $k_\lambda = \frac{k}{|1-\lambda k|}$.
My attempt is: \begin{align*} \alpha_\lambda &= \alpha + \lambda n \\ \frac{d\alpha}{ds_\lambda}\frac{ds_\lambda}{dt} &= \frac{d\alpha}{ds}\frac{ds}{dt} + \lambda \frac{dn}{ds}\frac{ds}{dt} \\ \frac{d\alpha}{ds_\lambda}|\alpha_\lambda '| &= \mathbf{t}|\alpha '| - \lambda k\mathbf{t}|\alpha '| \\ \frac{d\alpha}{ds_\lambda}|\alpha_\lambda '| &= \mathbf{t}|\alpha '|(1 - \lambda k). \end{align*}
Then by taking absolute value, we get $$ |\alpha_\lambda '| = |\alpha '||1 - \lambda k|. $$ This gets me the $|1 - \lambda k|$ term, but I have no idea how to get $k$ nor $k_\lambda$ from what I have left.