$$\sum_{n \leq N} e(n\alpha) \Lambda(n) =\left(\sum_{p \leq N} \log(p) \sum_{r=1}^{a(p): p^{a(p)} \leq N} e(p^r\alpha)\right)$$

But, the right hand side is gives, $$\left(\sum_{(a,q)=1 ,a<q} e(a\frac{p}{q})\sum_{\substack{n \leq N \\ n \equiv a (\text{modulo q})}} \Lambda(n) \right)=\sum_{\substack{p \leq N \\ p \nmid q}} \log(p) \sum_{r=0}^{a(p): p^{a(p)} \leq N} e(p^r\alpha)$$.

[ This is because $q_i \nmid n, \text{where} n \equiv a (\text{modulo q}), (a,q)=1$ ].

$\chi(q)=\sum_{\substack q_i} \log(q_i) \sum_{r=0}^{b_i: {q_i}^{b_i} \leq N} e({q_i}^r\alpha)$.

This, $\chi(q)$ obviously depends on the $N$.

$$\sum_{n \leq N} e(n\alpha) \Lambda(n) =\left(\sum_{p \leq N} \log(p) \sum_{r=1}^{a(p): p^{a(p)} \leq N} e(p^r\alpha)\right)$$

But, the right hand side is gives, $$\left(\sum_{(a,q)=1 ,a<q} e(a\frac{p}{q})\sum_{\substack{n \leq N \\ n \equiv a (\text{modulo q})}} \Lambda(n) \right)=\sum_{\substack{p \leq N \\ p \nmid q}} \log(p) \sum_{r=0}^{a(p): p^{a(p)} \leq N} e(p^r\alpha)$$.

[ This is because $q_i \nmid n, \text{where} n \equiv a (\text{modulo q}), (a,q)=1$ ].

$$\chi(q)=\sum_{\substack q_i} \log(q_i) \sum_{r=0}^{b_i: {q_i}^{b_i} \leq N} e({q_i}^r\alpha)$$.

($q_i$ s are prime factors of $q$)

This, $\chi(q)$ obviously depends on the $N$ and isn't zero generally.