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Source: arXiv v2, pp. 5--6, Lemma 3.3 and its proof.

Statement

Let $s=m(\mathcal A)$ and $1\le j\le J$.

(a) If $\delta_i=0$ for every $i<j$, then

$$M_j^{(1)}\le s \sum_{\substack{d\ge d_1\\P^+(d)=p_j}}\frac1d. \tag{1}$$

(b) Uniformly in all choices $0\le\delta_i\le1/2$,

$$M_j^{(2)}\ll\frac{s^2(\log p_j)^6}{p_j^2}, \tag{2}$$

with an absolute implied constant.

Full proof relative to the external distortion bound

Write $p=p_j$ and let $t\in\{1,2\}$. Raise the nonnegative inequality of Lemma 3.2 to the $t$th power, expand the ordered product, and take expectation. This gives

$$\mathbb E_{j-1}\alpha_j^t\le \sum_{\substack{1\le r_1,\ldots,r_t\le\nu_j\\ g_1,\ldots,g_t\mid Q_{j-1}}} \sum_{\substack{1\le i_1,\ldots,i_t\le n\\ d_{i_\ell}=g_\ell p^{r_\ell}\ (1\le\ell\le t)}} \frac{\mathbb P_{j-1}\!\left( \bigcap_{\ell=1}^t(a_{i_\ell}+g_\ell\mathbb Z)\right)} {p^{r_1+\cdots+r_t}}. \tag{3}$$

Every occurring $g_\ell p^{r_\ell}=d_{i_\ell}$ is at least $d_1$. Once $g_\ell,r_\ell$ are fixed, multiplicity gives at most $s^t$ ordered choices of the indices. A compatible intersection is one progression of modulus $[g_1,\ldots,g_t]$; an incompatible one is empty. Applying the exact external distortion bound to every nonempty intersection yields

$$\mathbb E_{j-1}\alpha_j^t\le s^t \sum_{\substack{1\le r_1,\ldots,r_t\le\nu_j\\ g_1,\ldots,g_t\mid Q_{j-1}\\ g_\ell p^{r_\ell}\ge d_1\ (1\le\ell\le t)}} \frac{\displaystyle\prod_{p_h\mid[g_1,\ldots,g_t]} (1-\delta_h)^{-1}} {[g_1,\ldots,g_t]p^{r_1+\cdots+r_t}}. \tag{4}$$

For $t=1$ and $\delta_h=0$ for $h<j$, the distortion product is $1$. Each pair $(g,r)$ determines $d=gp^r$ with $P^+(d)=p$, so (4) is bounded by the enlarged sum in (1).

For $t=2$, each distortion factor is at most $2$, whence

$$M_j^{(2)}\le\frac{s^2}{(p-1)^2} \sum_{g_1,g_2\mid Q_{j-1}} \frac{2^{\omega([g_1,g_2])}}{[g_1,g_2]}. \tag{5}$$

Indeed, the two geometric sums in $r_1,r_2$ are each at most $\sum_{r\ge1}p^{-r}=1/(p-1)$.

The remaining sum is multiplicative. If a prime power $q^e$ with $e\ge1$ is the exact $q$-part of $[g_1,g_2]$, there are

$$(e+1)^2-e^2=2e+1$$

ordered pairs of exponents having maximum $e$, and the factor $2^{\omega([g_1,g_2])}$ contributes $2$. Therefore

$$\sum_{g_1,g_2\mid Q_{j-1}} \frac{2^{\omega([g_1,g_2])}}{[g_1,g_2]} =\prod_{h<j}\left( 1+2\sum_{e=1}^{\nu_h}\frac{2e+1}{p_h^e}\right). \tag{6}$$

Each local factor is $1+6/p_h+O(p_h^{-2})$. Enlarging the absolute constant handles the finitely many small primes, and $1+z\le e^z$ gives

$$(6)\ll\exp\left(6\sum_{p_h<p}\frac1{p_h}\right) \ll(\log p)^6$$

by Mertens' estimate for the prime reciprocal sum. Finally $(p-1)^{-2}\ll p^{-2}$, and (2) follows. The Chinese remainder theorem and Mertens' estimate are standard external inputs; the distortion estimate is the separately identified imported lemma above.