Q.Show that $[(\sim q)\wedge p]\wedge q$ is a contradiction.

A truth table lays out, in every possible combination of $T/F$ for the component simple statements, the resulting truth value of a compound statement formula -- with $2^n$ rows for $n$ variables (2 for a 1-variable formula, 4 for two variables, 8 for three, and so on).

Building one systematically. List the variables as leading columns (all $T/F$ combinations, conventionally in a fixed order such as TT, TF, FT, FF for two variables), then add one column per sub-expression, working from the innermost connective outward, until the final column is the value of the whole formula.

Classifying a formula by its final column:

• Tautology ($\mathbb T$): the final column is $T$ in every row -- the formula is true no matter what truth values its parts take. Example: $p\vee\neg p$ (the Law of Excluded Middle) is a tautology.
• Contradiction ($\mathbb F$): the final column is $F$ in every row. Example: $p\wedge\neg p$ is a contradiction.
• Contingency: neither -- the final column is a genuine mix of $T$s and $F$s, so the formula's truth value really does depend on $p,q,\ldots$ …