The value of $K_p$ for the reaction
$$\ce{Br2(l) + Cl2(g) <=> 2BrCl(g)}$$
is $\pu{1 atm}$. At equilibrium in a closed container partial pressure of $\ce{BrCl}$ gas is $\pu{0.1 atm}$ and at this temperature the vapour pressure of $\ce{Br2(l)}$ is also $\pu{0.1 atm}$. Then what minimum moles of $\ce{Br2(l)}$ to be added to $\pu{1 mol}$ of $\ce{Cl2}$ initially to get the equilibrium mixture from above?
In this question should I include the vapour pressure of $\ce{Br2}$ in calculating $K_p$ or not?
Yes
As the reaction is written above, you may incorrectly conclude the bromine is in a liquid state throughout and hence is omitted from the equilbrium constants.
However as the question reminds us bromine is highly volatile and significant amounts of gas will accumulate above the liquid, hence most of the reactions will be between gas molecules of bromine and chlorine rather than gas and liquid.
The vapour pressure tells us the partial pressure of these gas. Liquid bromine will evaporate until the sealed vessel contains bromine vapour with a partial pressure of 0.1 atm.
This means it is actually the equilibrium constant for the all three gases: $$K_p=\frac{[\ce{BrCl}]^2}{[\ce{Br2}][\ce{Cl2}]}$$ where $[\ce{A}]$ is the partial pressure of $\ce{A}$. As the vapour pressure of $\ce{Br2}$ is known to be 0.1 atm, that is the partial pressure of bromine you use in your equilibrium constant.
Hence the minimum amount of bromine to add to establish any equilibrium with the liquid is the point where partial pressure of bromine is 0.1 atm and all the bromine is in the gas phase. Any more bromine added will condense out back into liquid.
