Using a truth table, is the sentence below a tautology, contradiction or contingency? (P ⇒ Q) ⇔ (¬P ∨ Q) Also, I am not sure what the double headed arrow is supposed to mean. I know a single headed arrow means "implies" but I am not sure about the double headed one.

If we setup a truth table:

$\begin{array}{|c|c|c|c|} \hline P & Q & \mathbf{P \rightarrow Q} & \mathbf{\
eg P \vee Q} \\\ \hline 0 & 0 & \mathbf{1} & \mathbf{1}\\\ \hline 0 & 1 & \mathbf{1} & \mathbf{1}\\\ \hline 1 & 0 & \mathbf{0} & \mathbf{0}\\\ \hline 1 & 1 & \mathbf{1} & \mathbf{1}\\\ \hline \end{array}$

$\begin{array}{|c|c|c|c|} \hline P \rightarrow Q & \
eg P \vee Q & \mathbf{(P \rightarrow Q) \leftrightarrow (\
eg P \vee Q)} \\\ \hline 1 & 1 & \mathbf{1} \\\ \hline 1 & 1 & \mathbf{1}\\\ \hline 0 & 0 & \mathbf{1}\\\ \hline 1 & 1 & \mathbf{1}\\\ \hline \end{array}$

Your expression gives nothing but $1$, so therefore it's a tautology.