What is this resistor for in this circuit?

The 10K and 1K resistors in the circuit form a voltage divider when the switch is pressed closed. With the +12V supply this divider nominally sets a transistor base bias voltage at about 1 volt. Very little base current flows due to the fact that the emitter of the NPN transistor is held above ground and as such the NPN base emitter voltage never gets high enough to let the transistor turn on. In a simulation of such circuit with a 2N3904 transistor model it shows that the presence of the 1K resistor keeps some bias across the LED of about 0.7V due to very low level currents in the transistor . If the 1K resistor is removed and when the switch is closed to GND then the bias across the LED drops to essentially zero because the transistor becomes fully turned off.

From a functional standpoint to get an LED to turn on and off from the switch there is no need to have the 1K resistor as relates to just this simple circuit. On the other hand if this circuit was used in a more complex system that had a monitor circuit across the LED looking for the above mentioned bias it could be an indicator that all the wiring from the switch to the LED was intact and in place. In an real burglar alarm system where the switch and LED may be located far apart this residual bias detection can play a role to ensure that the wiring has not been tampered with.

You are right, the 1 kΩ resistor is pointless. When the switch is closed, is causes the base of the transistor to go low enough to turn it off, but outright shorting the base to ground would unambiguously achieve the same effect.

I don't really like this circuit much. In this case, I don't see the point to putting the LED in the emitter leg. It seems a convoluted way of doing things with no real benefit.

Given all the above, I wouldn't look to anything in that book as examples of good design.

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.