$ ( \neg ( \eta \wedge  \eta )\wedge (p \wedge \neg p))\ rightarrow (\eta \vee p)\wedge \neg ((\eta \vee p)) $. And by POS 3 we have $ (\eta \vee p)\wedge \neg ((\eta \vee p))\rightarrow (\eta \vee p) $. So we obtain $ p\wedge \neg p  \vdash \neg (p \wedge \neg p) \vee p $