I think an easier way to do this is not to try and relate P and Q's times to each other, but to keep them with the same time t and instead relate their initial displacements.
So if we focus on just P in the first second when it is projected, we should calculate how far it moves in this time:
s = ut + 0.5at^2
s = 20 x 1 + 0.5 x -9.8 x 1^2
s = 7.1
We should also calculate P's final velocity after one second:
v = u + at
v = 20 + -9.8 x 1
v = 2.2
Now, at the instant Q is projected (one second after P), we know P's initial displacement from O (7.1 metres) and its velocity (2.2 metres per second upwards). We can set Q's initial displacement from O to be 0 metres per second, and its velocity to be 20 metres per second upwards. From here, both particles are in the same reference frame in terms of time t, so we can now just equate them via simultaneous equations:
s_P = s_Q
2.2t +0.5 x -9.8 x t^2 + 7.1 = 20t + 0.5 x -9.8 x t^2
2.2t + 7.1 = 20t
t = 0.40 seconds (2.s.f.)
Then of course you have to add 1 second to the answer for the time it took Q to be projected in the first place.
So the final final solution is
t = 1.40 secondsShown on Wolfram Alpha below:

Hope this helps! Let me know if you need help with part b) as well.
