does this mean both p and q are initial states?

Question

Answer 1

When we describe state sets by propositional logic formulas, i.e., a symbolic encoding, then the represented states correspond with the satisfying assignments of the formula. In this case, we will have three initial states as you can see by the truth table:

P	Q	TRUE<->P|Q
0	0	0
0	1	1
1	0	1
1	1	1

 

So, initial states are {p}, {q}, and also {p,q}. Having said that, it don't recommend truth tables to find the satisfying assignments. We have better options using e.g. BDDs. 

Answer 2

First of all I would transform true <->q|p, this is equivalent to q|p.

Now, every state satifying q|p is an initial state. The states satisfying q|p are {p}, {q}, {p,q}.

The variables not used in the sets are negated, this means {p} is equal to p&!q