Why won't this PNP Transistor trigger?
Even if Q6 could be turned on completely, which is not what happens in this circuit, it will have a \$V_{BE}\$ drop \$\approx 0.7V\$, hence on its emitter you would find \$3.3V-0.7V=2.6V\$ approximately. Therefore, even if \$V_{CE}\$ were almost zero (as I said, impossible in your circuit), Q8 base wouldn't be pulled to ground. 

Get rid of R22 from the emitter and put it in series with Q6 base to set a suitable bias for Q6 to be turned on. With this modified schematic, Q6 acts as a switch and can pull Q8 base very close to ground (not exactly, however: you will have the smallish saturation voltage \$V_{CE(sat)}\$ across CE terminals of Q8, less than 200mV probably).

As a side note, with R22 on the emitter, Q6 acts as a constant current source, with output current being its 
\$
I_C \approx I_E 
= \frac{V_B \;-\;0.7V}{R22} 
= \frac{3.3V \;-\;0.7V}{220\Omega}\approx 12mA
\$. 

The problem is that that circuit works as a current source as long as it has voltage headroom towards the 12V rail. In your circuit it forces those 12mA into R25 (2.2k&Omega;) in parallel with the BE junction of Q8 (assuming you connect Q8 correctly, i.e. you swap C and E in your circuit).

That current splits and flows almost entirely in the forward biased BE junction of Q8. Why? Because if that junction were OFF, the entire 12mA current would flow in R25, and this would require a 26V drop across it, which is not possible with a 12V rail. Hence the BE junction must be ON and as such it will show a ~0.7V drop across it, which will impose a very small current in R25 (\$\frac{0.7V}{2.2k\Omega}\approx 0.31mA &lt;&lt; 12mA\$).

A 12mA current in its base is more than enough to saturate the output transistor and make it work as a turned-on switch (which is what you need). However, you won't get its base pulled to ground, as you would expect, because the "driver" transistor Q6 doesn't work like a switch, but as a (switchable)  current source.