a, $P1$: PTHH : $2NaOH + H_2SO4$ $\rightarrow$ $Na_2SO_4 + H_2O$
$P2$: PTHH : $2KOH + H_2SO_4$ $\rightarrow$ $K_2SO_4 + H_2O$
b, $P1$ : $n_{H_2SO_4}$ = $\frac{196}{98}$ = $2 (mol)$
$n_{NaOH}$ = $\frac{60}{40}$ = $1,5 (mol)$
$P2$ : $n_{NaOH}$ + $n_{KOH}$ = $2n_{H_2SO_4}$
$\Rightarrow$ $1,5$ + $n_{KOH}$ = $2.2$
$\Rightarrow$ $n_{KOH}$ = $2.2 - 1,5$ = $2,5 (mol)$
$m_{KOH}$ = $2,5.56$ = $140 (g)$
$m_{ddKOH}$ = $\frac{140.100}{40}$ = $350 (g)$