$\mathrm{PhC}\equiv \mathrm{CH}\frac{\text{ Piperidine }}{{\mathbf{I}}_2}\to \text{ Product }\left(\mathrm{A}\right)$

Product (A) is:

Piperidine 

is a strong base so it can abs tract terminal H-atom.

$\mathrm{PhC}\equiv \mathrm{C}-\mathrm{H};\stackrel{\text{ Piperdine }}{⟶}\mathrm{Ph}-\mathrm{C}\equiv {\mathrm{C}}^{\ominus }\stackrel{{\mathrm{I}}_2}{⟶}$

$$\begin{gathered} \mathrm{PhC}\equiv \mathrm{C}-\mathrm{I} \\ \left(\mathrm{A}\right) \end{gathered}$$