They're to prevent the transistors to conduct because of leakage current into the \$P-\$ and \$S-\$ outputs when they're supposed to be off. Any leakage current will go through the resistors and not the emitter-base junction as long as the voltage drop it causes is less than 0.7V.
If it were just the open-drain outputs and their pull-ups to define an output level the leakage would most likely be negligible. But now this current will be amplified by the transistor's \$H_{FE}\$, and that may be too high to be ignored.
edit
At high temperatures the output leakage current can go to almost a \$\mu\$A. The transistor may amplify that to more than 100\$\mu\$A from your supply, which you don't want. If we pick a 10k\$\Omega\$ for R2 the leakage current will cause a voltage drop across it (and thus across the transistor's base-emitter junction) of maximum 10mV, which is way too low to get the transistor conducting.
Calculation of R1:
If the main supply is 3V and your equipment needs 100mA you'll need at least 1mA of base current (many small signal transistors have an \$H_{FE}\$ of at least 100). Let's play safe and go for 5mA base current. Don't forget to check the current through R2: 0.7V/10k\$\Omega\$ = 70\$\mu\$A. OK, that's only 1% so we'll ignore that. Then R1 = (3V - 0.7V)/5mA = 460\$\Omega\$. (Yes, I know, I used other values in my comments, but with the given assumptions these should be better values).
