OK, here is what is going on. First, a LED is a current device so hold that thought. Now inside the SSR is a LED and in this case according to the data sheet that LED has a forward working voltage of 1.2 volts which is pretty typical and a forward working current of 20 mA or .020 Amp. If we were to directly apply 5 volts across the LED it would draw much greater current and burn up. So we need to limit the LED current. In this case limit that current to 20 mA. We know the LED forward voltage drop is 1.2 volts so if we apply 5 volts we can say 5.0 - 1.2 = 3.8 Volts. Therefore I can say 3.8 Volts / .020 Amp = 190 Ohms. This all goes back to Ohms Law.
So now if we look at the circuit I posted above if resistor R1 is dropping 3.8 Volts and the SSR internal LED is dropping 1.2 Volts there is our 5.0 volts and the circuit current will be 20 mA. The small problem is that 190 Ohms is not a common off the shelf resistor. Common off the shelf values would be 180 Ohms or 200 Ohms. We could get by using either value. Generally the next higher value is selected so we would use 200 Ohms. Doing so will slightly lower the current but not enough to likely matter.
Yes, the current draw off the 5 volt pin or 5 volt supply will be limited to 20 mA.
Ron