Question:

<p><strong>Step 1: Set up the truth table.</strong> We have two propositional variables p and q, giving 4 cases to evaluate.</p><p><strong>Step 2: Evaluate for p = T, q = T</strong><br/>• $p \to q = T \to T = T$<br/>• $\sim p = F$, so $\sim p \to q = F \to T = T$<br/>• $(\sim p \to q) \to q = T \to T = T$<br/>• $(p \to q) \to [(\sim p \to q) \to q] = T \to T = T$ ✓</p><p><strong>Step 3: Evaluate for p = T, q = F</strong><br/>• $p \to q = T \to F = F$<br/>• $\sim p = F$, so $\sim p \to q = F \to F = T$<br/>• $(\sim p \to q) \to q = T \to F = F$<br/>• $(p \to q) \to [(\sim p \to q) \to q] = F \to F = T$ ✓</p><p><strong>Step 4: Evaluate for p = F, q = T</strong><br/>• $p \to q = F \to T = T$<br/>• $\sim p = T$, so $\sim p \to q = T \to T = T$<br/>• $(\sim p \to q) \to q = T \to T = T$<br/>• $(p \to q) \to [(\sim p \to q) \to q] = T \to T = T$ ✓</p><p><strong>Step 5: Evaluate for p = F, q = F</strong><br/>• $p \to q = F \to F = T$<br/>• $\sim p = T$, so $\sim p \to q = T \to F = F$<br/>• $(\sim p \to q) \to q = F \to F = T$<br/>• $(p \to q) \to [(\sim p \to q) \to q] = T \to T = T$ ✓</p><p><strong>Step 6: Conclusion.</strong> In all four cases, the statement evaluates to T. Therefore, the given statement is a <strong>tautology</strong>.</p><p><strong>∴ Answer: B</strong></p>