The shear stress acts on a single surface. (If the load $\mathrm{P}$ was large enough to break the pin, it would break on only one surface.) The area subjected to shear stress in this instance is simply the cross-sectional area of the pin.

${\text{τ}}_{\mathrm{pin}}=\frac{\mathrm{P}}{{\mathrm{A}}_{\mathrm{shear}}}=\frac{\mathrm{P}}{{\mathrm{A}}_{\mathrm{pin}}}$