2. Connect $A K$.
$$\begin{array}{l} \text { Given } S T / / P Q \\ \Rightarrow \frac{S K}{K Q}=\frac{T K}{K P} . \end{array}$$

Notice, $\frac{S K}{K Q}=\frac{S_{\triangle A S K}}{S_{\triangle A K Q}}=\frac{A S \sin \angle S A K}{A Q \sin \angle K A Q}$.
Since $\angle S A K=\angle T C K, \angle K A Q=\angle P C K$, then
$$\frac{S K}{K Q}=\frac{A S \sin \angle T C K}{A Q \sin \angle P C K} \text {. }$$

Notice, $\frac{T K}{K P}=\frac{S_{\triangle C T K}}{S_{\triangle C K P}}=\frac{C T \sin \angle T C K}{C P \sin \angle P C K}$. Combining (1) and (2), we get $\frac{A S}{A Q}=\frac{C T}{C P}$.
$$\begin{array}{l} \text { Also, } \angle S A Q=180^{\circ}-\angle B A D \\ =180^{\circ}-\angle B C D=\angle T C P \\ \Rightarrow \triangle A S Q \backsim \triangle C T P \\ \Rightarrow \angle A S K=\angle C T K . \end{array}$$

Therefore, $B 、 S 、 K 、 T$ are concyclic.