As gnometorule said, your extra term is simply redundant.
Informally, note that if $p\land q$ is true, then both $p$ and $q$ are true, so $p\lor q$ is certainly true; thus, allowing the extra possibility $p\land q$ in addition to $p\lor q$ adds nothing. More formally,
$$\begin{align*} (p\lor q)\lor(p\land q)&\equiv p\lor\Big(q\lor(p\land q)\Big)&&\text{associativity}\\\ &\equiv p\lor\Big(q\lor(q\land p)\Big)&&\text{commutativity}\\\ &\equiv p\lor q&&\text{absorption}\;. \end{align*}$$