The pad is giving 6dB of power attenuation. Assuming a source resistance of 75 ohms and 1V open-circuit source voltage, the voltage across the pad input is 0.5V across 75 ohms for a power of
$$p_{in} = \frac{(0.5)^2}{75} = 3.33mW$$
The output voltage is .2047V and this is across 50 ohms so the output power is
$$p_{out} = \frac{(0.2047)^2}{50} = 0.838mW$$
The power is thus attenuated by
$$-10 \cdot \log{\frac{0.838}{3.33}} = 6\mathrm{dB}$$
I also tried simulation the 10dB balanced pi attenuator shown in the
page and get near perfect result from it.
If the source and load impedances are equal, the power and voltage attenuation are equal.
$$20 \log \frac{V_{out}}{V_{in}} = 10 \log \frac{V^2_{out}}{V^2_{in}} = 10 \log \frac{V^2_{out}}{Z_{out}}\frac{Z_{in}}{V^2_{in}} = 10 \log \frac{p_{out}}{p_{in}}, \, Z_{out} = Z_{in}$$
If the source and load impedances are unequal, the power and voltage attenuation will not be equal.
$$20 \log \frac{V_{out}}{V_{in}} = 10 \log \frac{p_{out}}{p_{in}} - 10 \log \frac{Z_{in}}{Z_{out}}$$
So, in your example, we should have
$$20 \log \frac{V_{out}}{V_{in}} = -6\mathrm{dB} - 10 \log \frac{75}{50} = -6\mathrm{dB} - -1.76 \mathrm{dB} = -7.76 \mathrm{dB}$$
which agrees with your results.
