On can use Linear Programming (LP) duality.
Consider the LP problem 
$$\beta':=\max \langle a',x\rangle : x\in P.\tag{P}$$ 
So $\beta'$ is minimal number s.t. $\langle a',x\rangle\leq\beta'$ for all $x\in P$.
Thus $\beta'\leq b'$, for $b'$ as in the question.

The [dual](https://en.wikipedia.org/wiki/Dual_linear_program#Vector_formulations) of (P) is 
$$
\beta^*:=\min \langle\lambda,b\rangle : \lambda\geq 0, \lambda^\top A=a'. \tag{D}
$$
So we see that (D) encodes all the possible $\lambda$ giving $\lambda^\top A=a'$.




[Strong duality](https://en.wikipedia.org/wiki/Dual_linear_program#Strong_duality) says that $\beta'=\beta^*$, i.e. there exists feasible $\lambda$ s.t.
$\langle \lambda, b\rangle=\beta'\leq b'$, as required. 

Strong duality is not so easy to show, and it appears to be equivalent to the question asked. (Note that easier to show weak duality, which is $\beta^*\geq\beta'$, does not give the needed inequality).