Well the first thing to do is to simplify the expression.
$(p \wedge q) \rightarrow (p \vee q)$
By material implication
$\neg (p \wedge q) \vee (p \vee q)$
By De Morgan's law
$(\neg p) \vee (\neg q) \vee (p \vee q)$
$((\neg p) \vee p) \vee ((\neg q) \vee q)$
Which is always true because $((\neg p) \vee p)$ is always true and $((\neg q) \vee q)$ is also always true.
So the logic table contains all T regardless of what p and q are assigned to.
Alternatively, you can painfully substitute, which I will start here:
$p = T$, $q = T$ gives:
$(T \wedge T) \rightarrow (T \vee T)$
$(T \wedge T) = T$, $(T \vee T) = T$, and $T \rightarrow T = T$
so this case is T.
You can do likewise for the other 3 cases.