Show that (p ∧ q) → (p ∨ q) is a tautology?

It is because of the following equivalence law, which you can prove from a truth table: $$r\rightarrow s\equiv \lnot r\lor s.$$ If you let $r = p\land q$ and $s = p\lor q$, you get what you are looking for, namely that $$(p\land q)\rightarrow (p\lor q)\equiv \lnot(p\land q)\lor(p\lor q).$$

This is a classic equivalence $$p\to q \equiv \neg p\lor q.\tag 1$$

We can examine equivalence using a truth table \begin{array}{c|c|c|c|c} p&q&\neg p&p\to q&\neg p\lor q\\\hline T&T&F&T&T\\\hline T&F&F&F&F\\\hline F&T&T&T&T\\\hline F&F&T&T&T \end{array}

Hence, $p\to q\equiv \neg p\lor q$.

Further, if $a \equiv p\land q$ and $b\equiv p\lor q$, and so
$$a\to b\equiv \neg a\lor b\equiv \neg(p\land q)\lor(p\lor q)\equiv (\neg p\lor \neg q)\lor (p\lor q),$$ by using $(1)$.

To show (p ∧ q) → (p ∨ q).

If (p ∧ q) is true, then both p and q are true, so (p ∨ q) is true, and $T \to T$ is true.

If (p ∧ q) is false, then (p ∧ q) → (p ∨ q) is true, because false implies anything.

Q.E.D.

I know you asked specifically about a given proof, but here is another way:

(1) Assume p ∧ q

(2) By ∧-elimination, p

(3) By ∨-introduction, p ∨ q

(4) By →-introduction and marking the assumption (1), (p ∧ q) → (p ∨ q).

In less formal language: if P and Q is true, then you can look at either P or Q separately and it must be true. Now from any true statement you can create a longer true statement by creating a disjunction with any statement: if P is true, then "P or R" is also true for any statement R (e.g. if you are sure "it is raining", then it is also the case that "it is raining or you are a dragon"). In particular taking R = Q in this case allows you to reason that P or Q is true.