For most practical purposes we can consider KOH as the strongest base in water, and so the highest pH should be that of its saturated solution. Not accounting for the activity coefficient effect, we can approximately estimate the pH of a saturated solution of KOH as 14.8. This figure should be taken as approximate because the pKa (or pKb) published data for strong acids or bases dissolved in water typically suffers from considerable uncertainty; particularly while considering its applicability for concentrated solutions.

The reasoning could be as follows ― to avoid the assumption of complete dissociation, somewhat coarse for the rather concentrated saturated solution:

The pH of KOH aq. sol. can be predicted from the hydrolysis constant (Kh = 3.2·10-15; units omitted) (*) of K+ for the equilibrium: K+ + 2 H2O ⇌ K(OH) + H3O+; Kh = [K(OH)]·[H3O+]/[K+]. We can substitute [K(OH)] at previous equation based on charge balance and molar balance for potassium, respectively: [K+] = [OH-] - [H3O+] = Kw/[H3O+] - [H3O+], [KOH] = CKOH - [K+] = CKOH - [OH-] + [H3O+] = CKOH - Kw/[H3O+] + [H3O+], where Kw is autodissociation constant of water (Kw = 1.0·10-14; units omitted) and CKOH is the nominal molar concentration of dissolved KOH (formality, to be precise). Hence we can write: Kh = {CKOH - Kw/[H3O+] + [H3O+]}·[H3O+]/{Kw/[H3O+] - [H3O+]} = {{[H3O+] + CKOH}·[H3O+] - Kw}·[H3O+] /{Kw - [H3O+]2}. As long as CKOH > 100·[H3O+] and CKOH > 100·Kh, this simplifies to: [H3O+]2 - (Kw /CKOH)·[H3O+] ≈ Kh·Kw/CKOH. This yields:

[H3O+] ≈ (½)·(Kw/CKOH)·{1 + √((4·CKOH/Kw) + 1/Kh)}·√Kh

With CKOH = 11.7 M (saturated aq. sol.): [H3O+] ≈ (½)·(1.0·10-14/11.7)·{1 + √((4·11.7/(1.0·10-14)) + 1/(3.2·10-15))}·√(3.2·10-15) = 1.71·10-15 M.

Hence: pH = - log10(1.71·10-15 M/M) = 14.8

(*) pKa = - log10(Kh) = 14.5: https://www.wou.edu/las/physci/ch412/Table2.htm