Let me remind some of notations for readers:
$$\mathcal{L}=\log x,$$
$$k_0=3.5\times 10^6, \varpi=\frac{1}{1168},$$
$$D_0=\exp(\mathcal{L}^{\frac{1}{k_0}}), P_0=\prod_{p < D_0}p,$$
$$D_1=x^{\varpi},P=\prod_{p < D_1}p,$$

$$D=x^{\frac{1}{4}+\varpi}, D_2 = x^{\frac{1}{2}-\epsilon}$$

Here, the effect of replacing $\theta$ by $\Lambda$ produces an error of
$$O(x^{\frac{1}{2}}\mathcal{L}^B),$$
for some positive $B$. So, this change is negligible compared to $O(x\mathcal{L}^{-A})$.

For imposing $(d,P_0)< D_1$, we claim that the following sum is also negligible compared to $O(x\mathcal{L}^{-A})$:
$$
\sum_{\substack{{D_2 < d {<} D^2} \\\ {d|P} \\\ {(d,P_0)\geq D_1}}} \sum_{c\in C_i(d)} |\Delta(\Lambda,d,c)|
$$
Note that the trivial bound for $|\Delta(\Lambda,d,c)|$ is:
$$
|\Delta(\Lambda,d,c)|=O(\frac{x\mathcal{L}}{d})$$

Since we have $(d,P_0)\geq D_1$, the number $w(d)$ of distinct prime divisors of $d$, must satisfy
$$
w(d )\geq \mathcal{L}^{\varpi(1-\frac{1}{k_0})}.$$

The sum is bounded by:
$$
x\mathcal{L} \sum_{\substack{{ d {<} D^2} \\\ {w(d )\geq \mathcal{L}^{\varpi(1-\frac{1}{k_0})} } }} \frac{\tau_{k_0}(d)}{d }$$

Standard argument now applies, and the sum above is bounded by:
$$
\frac{1}{2^{\mathcal{L}^{\varpi(1-\frac{1}{k_0})}}}\sum_{d < D^2} \frac{2^{w(d)}\tau_{k_0}(d)}{d}=O(\frac{\mathcal{L}^B }{2^{\mathcal{L}^{\varpi(1-\frac{1}{k_0})}}}),$$
for some positive constant $B$.

Combining all together, we obtain an upper bound:
$$
\sum_{\substack{{D_2 < d {<} D^2} \\\ {d|P} \\\ {(d,P_0)\geq D_1}}} \sum_{c\in C_i(d)} |\Delta(\Lambda,d,c)| =O(\frac{x\mathcal{L}^{B+1}}{2^{\mathcal{L}^{\varpi(1-\frac{1}{k_0})}} })$$

Hence, our claim follows.