I cannot see how the first inclusion in this proof works.
$P$ is the maximum number of disjoint $B(\epsilon/2)$ with centres in $A$ and the following will help.
Moreover I cannot see how it leads to $\overline{\text{dim}}_pA \leq d$. I believe that the definition of the upper Minkowski dimension may help, but I was unable to see how it would.
Instead of the definition of the upper Minkowski dimension, use the formula from page 78 (this is Mattila's book of course): $$ \overline{\dim}_M B =\limsup_{\epsilon\to0}\frac{\log P(B,\epsilon)}{\log(1/\epsilon)} \tag1 $$ As you recalled, the definition of $P(B,\epsilon)$ is that it's the maximal number of disjoint balls of radius $\epsilon$ with centers at the points of $A$. Since each ball has diameter at most twice the radius, it follows that $$P(B,\epsilon/2)\epsilon^s \le P^s_\epsilon(B) \tag2$$ (The thing on the right is supremum over certain packings, of which the packing by $\epsilon/2$-radius balls is one.)
Suppose $\mathcal P^s(A)=0$. This means we can write $A=\bigcup A_n$ with $\sum P^s(A_n)$ arbitrarily small. Fix $n$. Since $P^s(A_n)$ is finite, there is $\delta>0$ such that $P^s_\delta (A_n)<\infty$. Then for $0<\epsilon\le \delta$ we have $P^s_\epsilon (A_n)\le P^s_\delta (A_n)$, which by (2) implies $$P(A_n,\epsilon) = O(\epsilon^{-s}),\quad \epsilon\to 0$$ From (1) we get $\overline{\dim}_M A_n\le s$. Since this holds for every $n$, it follows that $\overline{\dim}_p A\le s$.