I suggest to buy $\ce{KCl}$, or to prepare it by neutralization of $\ce{KOH}$ and $\ce{HCl}$.
If it happens you have mixed solution of $\ce{NaCl}$ and $\ce{KOH}$, neutralize it first by $\ce{HCl}$ (pH meter or pH indicator paper strips).
From Wikipedia solubility table one can conclude that while $\ce{NaCl}$ solubility almost does not depend on temperature, solubility of $\ce{KCl}$ is almost twice at 100 °C, compared 0 °C.
My suggestion is fractional crystallization:
Partially evaporate the solution until significant mass of salts crystallizes.
Heat the mixture to near boiling and mix well - part of $\ce{KCl}$ would dissolve.
Separate the precipitate enriched by $\ce{NaCl}$.
Cool down solution and let crystalyze salt enriched by $\ce{KCl}$.
With both precipitates and residual solution, you can recursively repeat the procedure.
You can set a system of N batches, where $\ce{NaCl}$ enriched precipitate would go to direction of the batch more enriched by $\ce{NaCl}$ and vice versa.
If it looks complicated to you, buy $\ce{KCl}$ or remember the Pierre and Marie Curie struggles with the isolation of traces of $\ce{RaCl2}$ from $\ce{BaCl2}$, where their solubilities are much closer.
Another option is using about 3 times better solubility of $\ce{NaCl}$ in methanol, compared to $\ce{KCl}$. But as it is like $\pu{5 g/L}\ \ce{KCl}$ and $\pu{15 g/L}\ \ce{NaCl}$, there would be extensive solvent manipulation with price and safety issues.
As useful info, here is the question/answer about the solubility diagram of mixtures of both salts