Meh, it's a simple thing.
Ignore the stupid capacitor, it's an open circuit anyway. The diode and PNP are a current mirror, but there is 22kΩ in between which drops a bit of voltage due to PNP base current, so PNP collector current may actually be less than 10kΩ resistor current. Anyway, that gets multiplied by NPN β and causes some voltage to appear across the LEDs and 47Ω.
Now, the stupid capacitor. So NPN collector voltage is going to be noisy as hell and this couples through the cap to PNP base. Any negative edge at NPN collector pulls PNP base down, pushing more current into PNP and then NPN. Positive feedback occurs and the NPN quickly turns on, drawing extra PNP base current through 220kΩ and the cap. Until the cap charges, because then the extra PNP base current disappears, NPN collector voltage goes up and the same feedback process turns off both transistors very fast.
Then the cap discharges through 22kΩ. Rinse, repeat.