When excess of NaOH solution is added in potash alum the product is

Formula of potash alum is ${\mathrm{K}}_2{\mathrm{SO}}_4\cdot {\mathrm{Al}}_2{\left({\mathrm{SO}}_4\right)}_3\cdot 24{\mathrm{H}}_2\mathrm{O}$. Potash alum consists of ${\mathrm{Al}}_2{\left({\mathrm{SO}}_4\right)}_3$ which initially forms $\mathrm{Al}{\left(\mathrm{OH}\right)}_3$, when reacts with NaOH. As NaOH is in excess, therefore $\mathrm{Al}{\left(\mathrm{OH}\right)}_3$ further react with NaOH to form NaAlO₂ which is soluble and a clear solution is obtained.
$\begin{array}{l}{\mathrm{Al}}_2{\left({\mathrm{SO}}_4\right)}_3+6\mathrm{NaOH}⟶2\mathrm{Al}(\mathrm{OH}{)}_3+3{\mathrm{Na}}_2{\mathrm{SO}}_4\\ \mathrm{Al}(\mathrm{OH}{)}_3+\mathrm{NaOH}⟶\underset{\mathrm{Soluble}}{{\mathrm{NaAlO}}_2}+2{\mathrm{H}}_2\mathrm{O}\end{array}$