Sum of two polyhedra is a polyhedron

a)

Let $P=\{x|Ax\ge a\}, Q=\{y|By\ge b\}$.

Now define $M=\{(x,y,z)|Ax\ge a, By \ge b, z=x+y\}$.

$P+Q$ is the projection of $M$ on the $z$ coordinates, therefore a polyhedron.

b)

We want to show that $x$ must be an extreme point in $P$, if $z=x+y$ is an extreme point in $P+Q$

First let's assume that $x$ is not an extreme point in $P$, i.e. there are $x_1,x_2\in P\backslash\{x\}$ and $0<\lambda<1$ s.t. $$\lambda x_1 + (1-\lambda) x_2 = x$$

Since $x_1, x_2\in P$, also $z_1:=x_1+y, z_2:=x_2+y\in P+Q$ (this is how $P+Q$ is defined).

As you might have guessed, $\lambda z_1 + (1-\lambda) z_2 = z$ (it's easy to check this, just replace the $z_1, z_2$ with their definition, expand and simplify).

But wait, we depicted $z$ as a linear combination of elements of $P+Q$
_{(to be formally correct: $0<\lambda<1,\;z_1,z_2\in (P+Q)\backslash\{z\},\;\lambda z_1+(1-\lambda)z_2 = z$)}
Hence $z$ cannot be an extreme point in $P+Q$.

So we showed if $x$ is not an extreme point in $P$, then $z=x+y$ cannot be an extreme point in $P+Q$. Apply the contraposition and you're done: If $z=x+y$ is an extreme point in $P+Q$ then $x$ is an extreme point in $P$.

Note that $x, P$ and $y, Q$ are mutually interchangable.