The pull-down and the light

Three things on one wire

Try the switch both ways, then set VCC to Nothing. The whole board is in that drawing. VCC goes to one side of the switch. The other side is SIGNAL, and SIGNAL goes three places: to your board's input, to GND through a 10 kΩ resistor, and to GND through the red LED and a 1 kΩ resistor.

Why it needs a pull-down

An input pin only looks at a voltage. It takes almost no current, so a pin connected to nothing has nothing to set its voltage, and it drifts. It picks up whatever is nearby, a hand, a mains cable, the wire next to it, and reads HIGH and LOW at random. That is called floating.

With the switch open, SIGNAL would float. The 10 kΩ resistor to GND stops it: it is the only thing connected, so it pulls SIGNAL to 0 V, and the pin reads LOW. A resistor that holds a line at 0 V like this is a pull-down.

When the switch closes, it connects SIGNAL straight to VCC. The switch wins easily, because a closed contact is a fraction of an ohm and the pull-down is 10,000. SIGNAL goes to VCC and reads HIGH. The pull-down now carries a small, steady current, 0.5 mA from 5 V, which is its price.

So on is HIGH and off is LOW. The silkscreen says ACTIVE HIGH for that reason, and your sketch sets the pin as a plain INPUT. There is no pull-up on this board. A block with a pull-up and the switch to GND would read the other way round, and some do; this one does not.

The light belongs to the switch

The LED's anode is on SIGNAL and its cathode goes to GND through 1 kΩ. It lights whenever SIGNAL is HIGH, which is whenever the switch is on. It is not connected to anything your sketch controls. That makes it a reliable way to see the switch's position, and a useless way to see whether your sketch noticed.

The LED is red, and its exact part is not recorded. A red 0805 LED drops about 2.0 V at these small currents, somewhere between 1.8 and 2.2 V, and the resistor sets what it gets:

These are worked out, not measured. From 3.3 V the LED is a little dimmer; the XL LED book works through why a 3.3 V supply leaves an LED less. With the switch off, the whole block takes nothing at all.

What it costs to get wrong

Set the pin as an output and the switch becomes a short. With the switch on, SIGNAL is VCC, and a pin driving LOW against it has only its own driver to limit the current. That is how a pin dies. This block is read, never driven.