If weak base $\mathrm{B}$, so equilibrium in its solution is,

$\mathrm{B}_{(\mathrm{aq})}+\mathrm{H}_{2} \mathrm{O}_{(l)}+\mathrm{BH}_{(\mathrm{aq})}^{+}+\mathrm{OH}_{(\mathrm{aq})}^{-} \quad \ldots \text { (i) }$

In above ionic equilibrium of base, the constant is $\mathrm{K}_{b}$ and $\mathrm{H}_{2} \mathrm{O}_{(t)}$ is taken as constant…

$\mathrm{K}_{b}=\frac{\left[\mathrm{BH}^{+}\right]\left[\mathrm{OH}^{-}\right]}{[\mathrm{B}]}$

If this expression is multiplied \& divided by $\left[\mathrm{H}^{+}\right]$,

$\mathrm{K}_{b}=\frac{\left[\mathrm{BH}^{+}\right]\left[\mathrm{OH}^{-}\right]\left[\mathrm{H}^{+}\right]}{[\mathrm{B}] \quad\left[\mathrm{H}^{+}\right]}=\frac{\left[\mathrm{OH}^{-}\right]\left[\mathrm{H}^{+}\right]\left[\mathrm{BH}^{+}\right]}{[\mathrm{B}]\left[\mathrm{H}^{+}\right]}$

In it $\left[\mathrm{OH}^{-}\right]\left[\mathrm{H}^{+}\right]=\mathrm{K}_{w}$ and $\frac{\left[\mathrm{BH}^{+}\right]}{[\mathrm{B}]\left[\mathrm{H}^{+}\right]}=\frac{1}{\mathrm{~K}_{a}}$

Because, $\mathrm{BH}^{+}$(acid) $+\mathrm{B}_{\text {(aq) }}+\mathrm{H}_{\text {(aq) }}^{+}$

So, $\mathrm{K}_{b}=\frac{\mathrm{K}_{w}}{\mathrm{~K}_{a}}$ and $\mathrm{K}_{w}=\left(\mathrm{K}_{a}\right)\left(\mathrm{K}_{b}\right)$

According to above,

$\mathrm{K}_{b} \times \mathrm{K}_{a}=\mathrm{K}_{w}=1 \times 10^{-14}$

taking log both the side

$\therefore\left(-\log \mathrm{K}_{b}\right)+\left(-\log \mathrm{K}_{a}\right)=-\log \mathrm{K}_{w}=\log \left(1 \times 10^{-14}\right)$

$\therefore \mathrm{pK}_{b}+\mathrm{pK}_{a}=\mathrm{pK}_{w}=+14 \quad \ldots \text { (Eq.-iii) }$