Let's call quinine, BN
.........BN + HOH ==> BNH^+ + OH^-
initial.0.39M.........0........0
change...-x...........x.........x
equil...0.39-x........x........x
Kb = (BNH^+)(OH^-)/(BN)
You have pKa of the conjugate acid, convert to pKb for the base, substitute into the Kb expression and solve for (OH^-) then convert that to pH.
%protonation = [(OH^-)/(BN)]*100 = ?