The PNP transistor, is common-emitter, switching on the high-side. That means the emitter is permanently tied to +6V. Given that the base-emitter potential difference must be less than 0.7V for this transistor to be switched off, it's clear that the base must have a potential of above +5.3V.
You are told that the 555 output, when high, is not guaranteed to rise closer than 1.7V from its positive supply potential, meaning that it's possible for that output to reach only 6.0V - 1.7V = +4.3V, instead of the full +6V.
This maximum of +4.3V is not high enough to switch off the PNP transistor, which we discovered requires above +5.3V at its base, to switch off. The 555 output is a full volt short of that!
The author's solution is to employ an LED, whose forward voltage is significantly more than 1V. When the 555 output is low, the part of the circuit that concerns us looks like this:
simulate this circuit – Schematic created using CircuitLab
We see that there will be about 24mA of base current, to switch on the transistor. This is possible because the LED D1 is forward biased, and permits current to flow.
Note that the base resistor R1 is smaller than the one used for the NPN transistor, because there's no diode in the NPN's base path. The diode used here, in the PNP base path, drops about 1.5V, and what remains across the resistor is therefore not as much as for the NPN base resistor. Presumably we want the same current in both bases, when the respective transistor is on, so the resistors must be scaled accordingly.
Now let's see what happens when the 555 output rises, say to +1V:
simulate this circuit
The thing to recognise here is that the voltage at the base, and the voltage at node A (the LED cathode) have not changed. That's because diodes and base-emitter junctions (which are technically also diodes) always clamp their voltages to some maximum. The voltage across the base-emitter junction cannot ever rise above 0.7V. The voltage across the LED cannot rise above 1.5V.
In other words, as long as current is flowing through them, the combined voltage across the base-emitter junction and the LED will be 0.7V + 1.5V = 2.2V. The only component here whose voltage is changing is resistor R1. Its voltage reduced by the same amount we raised the 555 output, because the voltage across the other two elements in the path, the LED and base-emitter junction, is clamped to a combined maximum of 2.2V.
If you raise the 555 output potential further still, eventually you will reach a point at which the voltage across R1 becomes zero, where Ohm's law tells you that no base current can be flowing, and the transistor will turn off. That potential is +3.8V. That's when the remaining 6V - 3.8V = 2.2V across the base-emitter junction and LED becomes insufficient to forward bias them.
The LED has effectively shifted the switching threshold down from +5.3V to +3.8V, and this new threshold is well within the output voltage range of the 555.
The reason the 555 can't output 6V is simply that the transistors inside it which are responsible for producing the output potential are not ideal, and in fact are quite weak. When "on", these transistors have significant collector-to-emitter resistance still, and drawing any amount of current through them (such as the 24mA base current this circuit requires) causes them to drop some voltage internally. Under no load, the output won't suffer nearly so badly, but as load current increases, output signal amplitude diminishes.
