Purpose of a two-transistor circuit
When "I/O select" is low, Q1 drives Q2 'on'. 

With I/O select at 0.4V, the emitter of Q1 will be at 1.1V and the base of Q2 at about 4.3V so the current through Q1 will be about 3.2V/150\$\Omega\$ = 21mA. About 10mA goes through R2 which leaves about 11mA for the base of Q2. 

Q2 will drive about 100mA to 500mA depending the acceptable voltage drop. 

R2 prevents leakage in Q1 from turning Q2 partially on, and greatly speeds the turn-off of Q2 when I/O select goes high. R2 determines the base current of Q2. 

Two transistors are used rather than one to get more gain, so the loading on the input is lower- it will be in the 100's of uA rather than tens of mA. 



I looked up and found the chip- which appears to have been an ancient 256x8 bipolar OTP (One Time Programmable) PROM (Programmable Read-Only Memory).. and they are interrupting the Vcc (an unconventional strategy from an unconventional Engineer- Woz) . It would have had a voltage specification and maximum Vcc current spec. A typical similar part is the DM74S471, which had a maximum Vcc current of 150mA, well within the range I mentioned- and commensurate with the limited working voltage range of the chip- it is not guaranteed to work if the voltage drop of Q2 is too large.  



Edit: To follow up on question by @HerrderElektronik, below is the simulated switching performance (voltage across a 50 ohm load) with R2 = 68 ohms (pink) and with R2 = 68K (green). Transistor types 2N4403 and 2SAR533, similar to original circuit. 



As you can see, R2 greatly reduces the turn-off time of Q2.