Solution:

$ f_c=4 \mathrm{kHz} \quad K=500 \Omega $

$ \begin{aligned} \\ &L=\frac{K}{\pi f_c}=\frac{500}{\pi \times 4 \times 10^3}=39.79 \mathrm{mH} \\\\ &C=\frac{1}{\pi f_c K}=\frac{1}{\pi \times 4 \times 10^3 \times 500}=0.16 \mathrm{\mu f} \\ \end{aligned} \\ $

The , $ T $ the section consists of an inductor $L / 2$ i.e $19.9 \mathrm{mH}$ in each series branch and a capacitor of $0.16 \mu f$ in the shunt branch

The $\pi$ section consist of an inductor of $39.79$ $\mathrm{mH}$ in series branch and a capacitor of $\mathrm{c} / 2$ i.é $0.08 \mathrm{\mu f}$ in each shunt branch.