Simplify, equivalent for (p ∨ ¬q) ∧ (¬p ∨ ¬q)

$$(p \lor \lnot q) \land (\lnot p \lor \lnot q) \iff (p \land \lnot p) \lor \lnot q \iff \lnot q.$$

Here's the corrected table:

$$\begin{array}{cc|cccc|c} P & Q & \neg P & \neg Q & P\lor\neg Q & \neg P\lor\neg Q & (P\lor \neg Q)\land(\neg P\lor \neg Q)\\\hline T&T&F&F&T&F&F\\ T&F&F&T&T&T&T\\ F&T&T&F&F&T&F\\ F&F&T&T&T&T&T \end{array}$$

From this, you can see that $(p \lor \neg q)\land(\neg p \lor \neg q) \iff (\neg q)$

This is actually an instance of the following logical equivalence principle:

Adjacency

$(p \land q) \lor (p \land \neg q) \Leftrightarrow p$

So, with this principle, you can immediately say that:

$(p \lor \neg q) \land (\neg p \lor \neg q) \Leftrightarrow \neg q$

When you do boolean algebra to simplify expressions, this situation comes up a lot, so I highly recommend remembering this equivalence principle!