How does this PNP transistor circuit work?
This circuit is just two PNP transistors that are set to operate as either on or off. On being like a closed switch and off being open circuit between the collector and emitter terminals. Here though one transistor is connected directly to collector of first transistor so it will actually invert the operation of the first transistor. When the unconnected end of the 100k resistor is connected to negative terminal of 7 (or 9V) battery, current will flow through the 100k to establish -0.7 across base emitter junction and this turn on Q1. When Q1 is on, it’s collector emitter junction has a voltage of about -0.2V and allows current to flow through LED1. Since Vce1 is only -0.2V, this isn’t sufficient to turn Q2 on and thus LED2 is off. 

When the 100k resistor is connected to ground, vbe1 is less than -0.7V and no base current flows, thus Q1 is off, therefore LED1 is off. However since Q1 is off, it’s collector emitter junction appears open circuit and voltage across LED1 and R4 are zero since no current flows or just a very very small current flows really.  This open circuit voltage across Q1 collector emitter junction is about -7V which can allow base current in Q1 to flow and turn on Q2 this turning on LED2.

To verify this, simulate the circuit in a spice circuit simulator.  Or better yet build this circuit on breadboard.