A solution using just words and definitions of the connectives:
Let $$ \alpha = \big((p\rightarrow q) \lor (p \rightarrow r)\big) \rightarrow (q\lor r), \;\;\;\; \beta = p \lor q \lor r$$
Suppose $\alpha $ is true. Then either $ \big((p\rightarrow q) \lor (p \rightarrow r)\big) $ is false, or $ \big((p\rightarrow q) \lor (p \rightarrow r)\big) $ is true and so is $(q \lor r)$. Suppose the former. Then $(p \rightarrow q)$ and $ (p \rightarrow r) $ must both be false which can only happen when $p$ is true and the statements $q, r$ are botth false. Since $p$ is true $\beta $ is true. Now suppose the latter. Then $(q \lor r)$ is true which means at least one of $q, r$ is true which renders $\beta $ true. Hence if $\alpha $ is true then $\beta$ is true $ ----(1)$
Now suppose $\beta$ is true. Then at least one of $p, q, r$ is true. If one of $q$ or $r$ is true then $\alpha$ is true since the consequent is true. So suppose not. Then $p$ must be true. It suffices to consider only the case when the antecedent of $\alpha$ is true. In this case one of $ (p \rightarrow q) $ or $ (p \rightarrow r) $ is true. Then at least one of $ q $ or $r$ must be true since $p$ is true and hence the consequent is also true. All cases are disposed of. $(q \lor r)$ is true which means at least one of $q, r$ is true which renders $\beta $ true. Hence if $\beta$ is true then $\alpha$ is true $ ----(2)$
From $(1)$ and $(2)$ we may conclude that $ \alpha $ is logically equivalent to $\beta$.