If the switch is open the base voltage is determined by the LED's forward voltage, for instance 2 V + 0.7 V = 3.7 V. Then the base current is (12 V - 3.7 V)/ 10 kΩ = 0.83 mA.

If you close the switch the current through the 10 kΩ resistor will be split to go partly through the 1 kΩ resistor, and partly into the base. We know the base needs 3.7 V before the transistor will begin conducting. To have 3.7 V there the current through the 1 kΩ will have to be 3.7 mA, due to Ohm's Law. So if the transistor would conduct, its base current will be 3.7 mA less than the current from the 12 V supply through the 10 kΩ resistor.

But we saw that that current won't be higher than 0.83 mA, so everything will go through the 1 kΩ and the transistor won't conduct at all. Since it doesn't conduct we can ignore it for now, and calculate the base voltage from the resistor divider:

$ V_B = {1 k\Omega}{1 k\Omega + 10 k\Omega} \times 12 V = 1.09 V, which indeed is lower than the required 3.7 V.

What if the 1 kΩ was omitted? Then the ground current would increase from 1.09 mA to 1.2 mA, that's all. That 0.1 mA difference won't break the bank, so you might as well omit it.

Frankly, I don't think this is a good circuit. You close the switch to switch the LED off, instead of on, which is, well, OK, but it means that when the LED is off you'll still have a current of 1.1 mA flowing, for nothing. It would be a better idea to place the switch on the 10 kΩ side. Admitted, it's function would be inversed (closing would switch the LED on), but you won't have a current with the LED off. In that case you still can add a resistor to ground, but its value should be much higher: a 4.5 kΩ will draw 0.83 mA at 3.7 V base voltage. That 0.83 mA was the current coming from the 12 V supply, so that's the point at which the transistor only just begins to conduct. So the value must be higher than that. A 100 kΩ value will draw 37 µA when the transistor conducts, so the base will get 830 µA - 83 µA = 750 µA. If you don't care about the 10 % loss you can place the resistor. You can also omit it there (not replacing it with a wire!), then the base will float when the switch is open. For a bipolar transistor that's not really a problem, especially since you would need a high 3.7 V to get it conducting, but for a MOSFET that resistor would be required.

If the switch is open the base voltage is determined by the LED's forward voltage, for instance 2 V + 0.7 V = 3.7 V. Then the base current is (12 V - 3.7 V)/ 10 kΩ = 0.83 mA.

If you close the switch the current through the 10 kΩ resistor will be split to go partly through the 1 kΩ resistor, and partly into the base. We know the base needs 3.7 V before the transistor will begin conducting. To have 3.7 V there the current through the 1 kΩ will have to be 3.7 mA, due to Ohm's Law. So if the transistor would conduct, its base current will be 3.7 mA less than the current from the 12 V supply through the 10 kΩ resistor.

But we saw that that current won't be higher than 0.83 mA, so everything will go through the 1 kΩ and the transistor won't conduct at all. Since it doesn't conduct we can ignore it for now, and calculate the base voltage from the resistor divider:

\$ V_B = \dfrac{1 k\Omega}{1 k\Omega + 10 k\Omega} \times 12 V = 1.09 V\$, 

which indeed is lower than the required 3.7 V.

What if the 1 kΩ was omitted? Then the ground current would increase from 1.09 mA to 1.2 mA, that's all. That 0.1 mA difference won't break the bank, so you might as well omit it.

Frankly, I don't think this is a good circuit. You close the switch to switch the LED off, instead of on, which is, well, OK, but it means that when the LED is off you'll still have a current of 1.1 mA flowing, for nothing. It would be a better idea to place the switch on the 10 kΩ side. Admitted, it's function would be inversed (closing would switch the LED on), but you won't have a current with the LED off. In that case you still can add a resistor to ground, but its value should be much higher: a 4.5 kΩ will draw 0.83 mA at 3.7 V base voltage. That 0.83 mA was the current coming from the 12 V supply, so that's the point at which the transistor only just begins to conduct. So the value must be higher than that. A 100 kΩ value will draw 37 µA when the transistor conducts, so the base will get 830 µA - 83 µA = 750 µA. If you don't care about the 10 % loss you can place the resistor. You can also omit it there (not replacing it with a wire!), then the base will float when the switch is open. For a bipolar transistor that's not really a problem, especially since you would need a high 3.7 V to get it conducting, but for a MOSFET that resistor would be required.