A thermocouple is made from two metals, Antimony and Bismuth. If one junction of the couple is kept hot and the other is kept cold then, an electric current will

The phenomenon involved is the Seebeck effect: when the two junctions of a dissimilar-metal loop are kept at different temperatures, an emf $$E$$ is produced, driving a steady current around the closed circuit.

The magnitude and sign of the emf are obtained from
$$E \;=\; \int_{T_c}^{T_h} \left(S_{A}-S_{B}\right)\,dT \;=\; \left(S_{A}-S_{B}\right)(T_h-T_c)$$
where $$S_{A}, S_{B}$$ are the Seebeck (thermo-electric) coefficients of the two metals and $$T_h \gt T_c$$.

Approximate room-temperature coefficients are
$$S_{Sb} \approx -40\,\mu\text{V K}^{-1},\qquad S_{Bi} \approx -72\,\mu\text{V K}^{-1}$$

Hence
$$S_{Sb}-S_{Bi} \;=\; (-40) - (-72) \;=\; +32\,\mu\text{V K}^{-1} \gt 0$$
so the emf is positive when the Antimony junction is hotter: $$E = +32\,(T_h-T_c)\,\mu\text{V}$$.

“Positive” here means that the Antimony wire is at higher electric potential than the Bismuth wire at the hot junction. Conventional current therefore leaves the Antimony wire, travels through the loop, and re-enters the Bismuth wire. Tracing the complete path shows that at the cold junction the current must go from the Antimony wire into the Bismuth wire.

Thus, when one junction is hot and the other cold, the current in the thermocouple flows
from Antimony to Bismuth at the cold junction.

Option A which is: flow from Antimony to Bismuth at the cold junction.