Two implications imply the negation of $p$

Direct reasoning. If $p$ is false, the antecedents $p\land q$ and $\bar p\land\bar q$ of (1) can be true only if $q=1$, but then (1) requires $\bar q=1$, false. More systematically:

• If $p=1$ and $q=1$: (1) is $1\Rightarrow0$, false.
• If $p=1$ and $q=0$: (2) is $1\Rightarrow0$, false.
• If $p=0$: both antecedents are always false, so both implications are vacuously true.

Hence both implications hold simultaneously iff $p=0$.

Truth table. Writing $\text{impl}(x,y)=\bar x\lor y$, one checks that (1) and (2) are both true only in the rows where $p=0$.

Conclusion. $p$ must be false.