Official solution
~ p \[ \wedge \] (p \[ \vee \] q) \[ \to \] q
\[ \equiv \] (~ p \[ \wedge \] p) \[ \vee \] (~ p \[ \wedge \] q) \[ \to \] q
\[ \equiv \] C \[ \vee \] (~ p \[ \wedge \] q) \[ \to \] q [C = contradiction]
\[ \equiv \] (~ p \[ \wedge \] q) \[ \to \] q
\[ \equiv \] ~ (~ p \[ \wedge \] q) \[ \vee \] q
\[ \equiv \] (p \[ \vee \] ~ q) \[ \vee \] q
\[ \equiv \] (p \[ \vee \] q) \[ \vee \] (~ q \[ \vee \] q)
\[ \equiv \] (p \[ \vee\] q) \[ \vee \] t
so ~ p \[ \wedge \] (p \[ \vee \] q) \[ \to \] q is a tautology