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 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 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).

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 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 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 it is easier to show weak duality, which is $\beta^*\geq\beta'$, and this does not give the needed inequality).