These 2 circuits are equivalent indeed, given the condition that the frequency is fixed. This is implied by the fact that you have a fixed impedance for the capacitor. The impedance will change with frequency, and once that happens these circuits are no longer equivalent.
I know seems counter intuitive, but just imagine for a minute that the capacitor is like a resistor that decreases value as the frequency increases, therefore, having a "resistor" in parallel with the 1k load resistor, will make an equivalent impedance smaller than impedances. For that reason, you can say that it's equivalent to having a smaller resistor, such as the 75 ohms, plus a series capacitor.
A way to understand why they are not equivalent for every frequency is to think what would happen if your frequency increases a lot. The impedance of your capacitor would become very small, and for the circuit on the left side of Image B, it would look almost like a short to ground, whereas on the right side it would still have the 75 ohms of the resistor in series. Therefore, you have to set the frequency you want the have the equivalent circuit and chose the capacitor to achieve the desired impedance for that specific value.
Try simulating the following 2 circuits. At 10 kHz these values of capacitance will give you similar ratio as given by the article values. The waveforms are the voltages at the load of each circuit, and you can see they are the same for both circuits.
So if you ignore practical factors that could add small disparities between the circuits in image B, such as intrinsic capacitances, resistances, inductances and mechanical factors, you can say that they are equivalent and you can actually use it in a practical design, if you know your frequency will be fixed and steady.