According to the buffer solution the $\mathrm{pOH}$ of solution is :-

$\mathrm{pOH}={\mathrm{pK}}_{\mathrm{b}} + \log \frac{\left[\mathrm{Salt}\right]}{\left[\mathrm{base}\right]}$

If the molar concentrations of base and its conjugate acid are same then $\left[\mathrm{Salt}\right]=\left[\mathrm{Base}\right]$

So, the $\mathrm{pOH}={\mathrm{pK}}_{\mathrm{b}} + \log 1={\mathrm{pK}}_{\mathrm{b}}+0={\mathrm{pK}}_{\mathrm{b}}$

Hence, if the molar concentrations of base and its conjugate acid are same, then $\mathrm{pOH}$ of the buffer solution is same as ${\mathrm{pK}}_{\mathrm{b}}$ of base.