Alkali metal hydroxides solubility in water increases down the group. Because the mechanism of dissolution is the same in alcohols, why the anomaly?
Your question has multiple aspects to be discussed:
Upon your observation; Solutes follow Like dissolve like principle to decide the extent to which they will get dissolved into solvent. It means more ionic solute will dissolve more in a polar solvent than a non-polar solvent and vice-versa.
KOH has a greater ionic character than NaOH and hence needs a more polar solvent to get dissolved as compared to the equivalent amount of NaOH needs to be. Since alcohols are very less polar, they may break bindings of NaOH effectively but are not able to break those of KOH to that extent. As a result KOH is less soluble in alcohols. This is also evident from the fact that KOH is more soluble in water as compared to NaOH because water is a polar solvent and hence it finds KOH more perfect match than NaOH.
As we all know solubility is a temperature-dependent property and does not possesses same slopes for all solutes. Hence you may get a lower solubility for KOH at some temperature.
By the word "Alcohols" we do not have a specific solvent and hence different solute-solvent pairs are possible and do not know exactly which alcohol you're discussing as some halogenated alcohol have greater affinity for NaOH.
Because of smaller size of sodium ion, hydration ease is another factor you should not ignore while concerning this.
Finally there are several other factors which leads to the greater solubility which include charge density, bond stability, percent ionic character etc. studying which may or may not give results contrary to your question.