With $C=\lVert a\rVert_{p,\omega}'$ there holds $a_j^*\le j^{-1/p}C$. Hence, $$\frac{\lVert a\rVert_{p,\omega}}C\le\sup_nn^{-1+1/p}\sum_{j=1}^nj^{-1/p}n^{1/p}\le\sup_nn^{-1+1/p}\Bigl(1+\int_1^nx^{-1/p}dx\Bigr)=\sup_nn^{-1+1/p}\frac1{1-1/p}n^{1-1/p}=\frac p{p-1}.$$