An AC source is connected to a capacitor $C$. Due to a decrease in its operating frequency:

Capacitive reactance $X_C=\frac{1}{2\pi fC}$.If frequency $f$ decreases, $X_C$ increases.The conduction current $I_c=\frac{V}{X_C}$. Since $X_C$ increases, $I_c$ decreases.According to Maxwell's theory, the displacement current $I_d$ is equal to the conduction current $I_c$ in the circuit.Therefore, the displacement current decreases.