We construct a truth table to evaluate both sides of the equivalence.
\begin{tabular}{|c|c|c|c|c|c|c|c|}
\hline
\(p\) & \(q\) & \(r\) & \(q \land r\) & \(p \lor (q \land r)\) & \(p \lor q\) & \(p \lor r\) & \((p \lor q) \land (p \lor r)\)
\hline
T & T & T & T & T & T & T & T
T & T & F & F & T & T & T & T
T & F & T & F & T & T & T & T
T & F & F & F & T & T & T & T
F & T & T & T & T & T & T & T
F & T & F & F & F & T & F & F
F & F & T & F & F & F & T & F
F & F & F & F & F & F & F & F
\hline
\end{tabular}
Since the truth values in the columns for \( p \lor (q \land r) \) and \( (p \lor q) \land (p \lor r) \) are identical for all possible truth values of \(p, q, \) and \(r\), the given equivalence (the distributive law) is verified.