Okay, I will do that in great detail below, and you can do many of these steps at once, of course. Instantiating all cases for p and q yields:

p q
0 0 (0 <-> a V b & Xp) ^ (0 <-> 0 V a & Xq)
0 1 (0 <-> a V b & Xp) ^ (1 <-> 0 V a & Xq)
1 0 (1 <-> a V b & Xp) ^ (0 <-> 1 V a & Xq)
1 1 (1 <-> a V b & Xp) ^ (1 <-> 1 V a & Xq)
and thus
p q
0 0 !(a V b & Xp) ^ !(0 V a & Xq)
0 1 !(a V b & Xp) ^ (0 V a & Xq)
1 0 (a V b & Xp) ^ !(1 V a & Xq)
1 1 (a V b & Xp) ^ (1 V a & Xq)
which simplifies to (note that x&phi is equivalent to x&(phi[x<-1]))
p q
0 0 !(a V b & Xp) ^ !(a & Xq)
0 1 !(a V b & Xp) ^ a & Xq
1 0 0
1 1 (a V b & Xp)
and further to
p q
0 0 !(a V b & Xp) ^ !(a & Xq)
0 1 !(1 V b & Xp) ^ a & Xq
1 0 0
1 1 (a V b & Xp)
which results in
p q
0 0 !(a V b & Xp) ^ !(a & Xq)
0 1 0
1 0 0
1 1 a V b & Xp

So, we have no outgoing transitions in states {p} and {q}. For state {p,q}, we have outgoing transitions for a to anyone of the four states, and for b to the states {p} and {p,q} since Xp must hold then.

The transitions of state {} are more difficult. We further simplify the transition relation and get the following

!(a V b & Xp) ^ !(a & Xq)
= !a & !(b & Xp) ^ !(a & Xq)
= !a & !(b & Xp) ^ !(0 & Xq)
= !a & !(b & Xp) ^ !(0)
= !a & !(b & Xp)
= !a & (!b | !Xp)
= !a & !b | !a & !Xp

Hence, we have transitions to anyone of the four states with !a&!b and to the states {} and {q} for input !a.