The P, Q, S at the bottom form an OR gate; there's a path if either input is 1. But it's ANDed with T, so that W = 0 if (T = 1) AND (P = 1 OR Q = 1 OR S = 1). The top part is just the De Morgan dual of this: W = 1 if (T = 0) OR (P = 0 AND Q = 0 AND S = 0).
So W = NOT (T AND (P OR Q OR S)).

The P, Q, S at the bottom look more complicated than they are. They're drawn as P OR (Q OR S) but that's the same as (P OR Q OR S).

The P, Q, S at the bottom form an OR gate; there's a path if either input is 1. But it's ANDed with T, so that W = 0 if (T = 1) AND (P = 1 OR Q = 1 OR S = 1). The top part is just the De Morgan dual of this: W = 1 if (T = 0) OR (P = 0 AND Q = 0 AND S = 0).
So W = NOT (T AND (P OR Q OR S)).

The P, Q, S at the bottom look more complicated than they are. They're drawn as P OR (Q OR S) but that's the same as (P OR Q OR S).

edit
Your truth table may be easier to interpret if you list the resp. inputs in binary counting order:

T P Q S
0 0 0 0
0 0 0 1
0 0 1 0
0 0 1 1
0 1 0 0
0 1 0 1
etc.

Some relationships between a certain input and the output may become more clear, in this case only in the bottom half of the table the output will be zero. The bottom half is when T = 1.

The P, Q, S at the bottom form an OR gate; there's a path if either input is 1. But it's ANDed with T, so that W = 0 if (T = 1) AND (P = 1 OR Q = 1 OR S = 1). The top part is just the De Morgan dual of this: W = 1 if (T = 0) OR (P = 0 AND Q = 0 AND S = 0).
So W = NOT (T AND (P OR Q OR S)).

The P, Q, S at the bottom form an OR gate; there's a path if either input is 1. But it's ANDed with T, so that W = 0 if (T = 1) AND (P = 1 OR Q = 1 OR S = 1). The top part is just the De Morgan dual of this: W = 1 if (T = 0) OR (P = 0 AND Q = 0 AND S = 0).
So W = NOT (T AND (P OR Q OR S)).

The P, Q, S at the bottom look more complicated than they are. They're drawn as P OR (Q OR S) but that's the same as (P OR Q OR S).