The answer is easy , 2! Prove: Find a rectangle covers P with area not bigger than 2, and chose M be the centre of that rectangle. Then Q is also covered by it, so the convex hull of P and Q is covered by rectangle,whose area not bigger than 2. Construction: A isosceles right triangle, need some discussion.

Actually, you proved $\alpha\le 2.$ But $2$ is not the minimum value. I believe $\alpha=\frac{4}{3}$ and this value is attained when $P$ is a triangle. The optimal point $M$ is the centroid of $P.$

Note that the problem is about "the convex hull of $P \cup Q$ ", not $P \cup Q$.

Sorry, my bad, I've read it as "the area of $P\cup Q$" not "the area of the convex hull of $P \cup Q$ ". But, could someone check the translation again. I belive it should be "the area of $P\cup Q$". Because, otherwise the problem is next to trivial, and I do not believe a problem like that to be given at a Chinese TST.

dgrozev wrote: Sorry, my bad, I've read it as "the area of $P\cup Q$" not "the area of the convex hull of $P \cup Q$ ". But, could someone check the translation again. I belive it should be "the area of $P\cup Q$". Because, otherwise the problem is next to trivial, and I do not believe a problem like that to be given at a Chinese TST. In fact, it's "the convex hull of $P \cup Q$".Besides, many problems in Chinese TST are trivial.

I think the problem is not at all trivial. The idea of considering such a rectangle appears quite ingenious to me (though, guessing that the answer is $2$ seems not tough because of case of a triangle).