C3 is the wrong way round.
T4 is turned on by R9 (although that might be a bit marginal as it relies on a very good gain on T4)
The microphone signal is amplified by T1 and T2.
When there is a big enough signal on the collector of T2, the voltage comes down, causing the other end of C3 to come down as well. That turns off T4.
The collector of T4 goes up, turning on the Triac, and T3.
T3 holds the left end of C3 low, holding T4 off until current through R9 has brought the voltage at the right end of C3 up to 0.7 V, where T4 turns on, the Triac turns off, T3 turns off and the circuit goes back to how it was.
It should take several minutes for C3 to be charged though R9.
Problems.
Apart from C3 being the wrong way round, a BC548B has a gain of 220 to 475 (
Data sheet here) so the 11 μA through R9 will only be amplified to about 2 - 5 mA, so the voltage on the collector of T4 won't ever come down enough to turn off the Triac or T3. You should probably use two transistors instead of T4 to improve the gain.
When T3 turns on, C3 will take the base voltage of T4 very negative, past the base-emitter breakdown voltage. While that probably doesn't do any harm, half the capacitors charge will be lost quickly, so the capacitor has to be larger than it would otherwise be.
There is a reasonable explanation where the circuit diagram came from:-
**broken link removed**