Step 1: Understanding the Concept
\(\text{sp}^{3}\text{d}\) hybridisation needs five electron domains around the central atom, giving a trigonal bipyramidal shape.
Step 2: Check the Options
In \(\text{PCl}_{5}\), P has five bond pairs and no lone pair, so it uses one 3s, three 3p and one 3d orbital.
Step 3: Why the other options fail
\(\text{SO}_{2}\) has three domains (\(\text{sp}^{2}\)). \(\text{BF}_{3}\) has three domains (\(\text{sp}^{2}\)). \(\text{CO}_{2}\) has two domains (sp).
Final Answer:
\(\text{PCl}_{5}\) shows \(\text{sp}^{3}\text{d}\) hybridisation.
\[ \boxed{\text{(A) }\text{PCl}_{5}} \]