I can't resist some comments. I find it highly unlikely that what you ask is possible. The question seems a bit odd in that it is so specific when more basic questions are open.
In the following some estimates are rough and corrections are welcome. Let $p_i$ be the $i$th prime and $P_t$ the product of the first $t$ primes. For example $p_{10000}=104729$ and $P_{10000} \approx e^{104392.2}.$ This illustrates the estimate $\ln{P_t} \approx p_t.$
Define $h(t)$ to be the smallest $N$ so that out of every $N$ consecutive integers at least one is relatively prime to $P_t$. Equivalently, it is the largest $N$ such that some $N-1$ consecutive integers each have a prime divisor from among the first $t$ primes. Equivalent to that is that $h(t)$ is the largest $N$ such that there is a choice of $t$ residues
$ k_{1t},\cdots ,k_{tt} $ with $\cup_1^t A_{it} \supset [1,N-1].$ where $A_{it}=k_{it}+np_i.$
According to this article $h(t) \gt (2e^{\gamma}+o(1))\frac{p_t\ln{p_t}\ln_3{p_t}}{ln_2^2{p_t}}$ where $\ln_j$ is the $j$-fold iterated logarithm. This lower bound seems reasonably close for $t \lt 50$ (the limit of knowledge at the time, and perhaps now). The best known upper bound is $h(t) \lt \ln^2(P_t).$ Using the estimate above that $\ln{P_t} \approx p_t$ we then get that $h(t) \lt p_t^2.$ I have seen it conjectured in a paper from 1976 that perhaps $h(t)=O(t^{1+\epsilon}).$ Note that $p_t \approx t\ln{t}=o(t^{1+\epsilon})$ I don't know how current that is. An off topic note is that sometimes one can find a longer interval of integers all enjoying a divisor from $p_1,\dots,p_{t-1},p_{t+1}$
With the notation above a weaker form of your question is
Is there an integer $M$ and system of residues $k_i$ such that for $A_i=k_i+np_i$ we have
$\cup A_{i}=\mathbb{N}$
With only finitely many exceptions, $k_i \gt \frac{p_i^2}{M}$
(I didn't get the motivation to require $p_i$ relatively prime to $M$. Was it just to forbid at least one prime?)
So the situation is that, as far as I know, no-one can say if $h(n) \gt \frac{p_t^2}{M}$ infinitely often, and there may even be reasons to doubt that. This is the case that you may completely change the residues each time you bring a new prime into play. In your case you want a single list of residues.Note that the choices which provide a good example for $h(n)$ may not be good for extending to $h(n+1)$.
$h(9)=39$ as shown by
$$2, 3, 2, 5, 2, 11, 2, 3, 2, 13, 2, 23, 2, 3, 2, 7, 2, 19, 2, 3, 2, 17, 2, 5, 2, 3, 2, 11, 2, 7, 2, 3, 2, 5, 2$$
$$ 13, 2, 3, 2, *, 2, *, 2, 3, 2, *, 2, *, 2, 3, 2, *, 2, 5, 2, 3, 2, 7, 2, *$$
This illustrates that $1+2n,2+3n,4+5n ,6+11n,10+13n,12+23n,16+7n,18+19n$ and $22+17n$ cover all the integers from $1$ to $67$ except for $40,42,46,48,52,60,66.$
The best we can do to extend this introducing the further primes $29,31,37,41,43$ is $h(10) \ge 42,h(11) \ge 46,h(12) \ge 48,h(13) \ge 52,h(14) \ge 60$ etc.The actual values are $46, 58, 66, 74, 90.$ But each requires a new selection of residues.
A simple proof is available as well. Pick p coprime to d and let t be such that td=1 mod p. Then, mod p, t times the arithmetic progression looks like a sequence of consecutive integers. Thus its length has to be less than p to avoid one of the terms being a multiple of p, which means the original progression also has to have fewer than p terms. So the collection of primes dividing a set of arbitrarily long arithmetic progressions must be infinite.
It has been noted by GH from MO that the above overlooks some subtleties; the following is
more inspired by Euclid, and should leave no doubt remaining.
Let A be a set of arbitrarily long nonconstant arithmetic progressions. Thus for any
integer m, we can find a member of A and extract from that (wlog) a positive increasing
arithmetic progression of the form a +kd where a and d are coprime and k goes from 0 to
m. Pick a finite set of primes M, set m equal to a large multiple of their product (so at
least twice the product of primes of M) and then choose from A a progression and derive
the progression a +kd with the properties above. I will show the existence of a prime divisor
which is not in M and divides a member of the progression.
Now d may share some factors with the product of M, but as (a,d)=1, none of the factors
of d will divide any of a +kd. To be safe, let us call M' that set of numbers in M which are
coprime to d, and set their product to m'. So (m',d)=1, d is a unit mod m', and we can look
at ta +tkd as k runs from 1 to m which is bigger than m'. As above, td=1 mod m'.
Modulo m', ta+tkd is ta+k, so ta+k is divisible by a factor of m' precisely when a+kd is. As a result,
there are k bigger than m' and at most m such that a+kd is coprime with m'. But a+kd is bigger than
m' and is coprime not just to m' but to the product of all primes in M. So it must have a prime factor
outside of M. So A "contains" more primes than found in any finite set of prime factors.
Clear enough?
Best Answer
No. In fact, for $p = 435052917615787$, this will absolutely be false, as the second homology group will not vanish.
Note that a 2-dimensional "hole" in your complex is a 3-simplex all of whose faces are 2-simplices in $K(p)$, i.e. 4 primes such that each 3 of them lie in some arithmetic progression.
Of course, you would suspect that such a thing is plausible, and with some effort construct an example. Here's why this number works:
Consider $p = 398936189798617$. Then, if $d = 2124513401010$, one sees that $p+d$ is not a prime, while $p+2d,p+3d,\ldots,p+15d$ are all primes. Here $p+15d = 435052917615787$ is the prime for which we consider the complex.
In fact, the numbers were taken from the AP-k records at http://primerecords.dk/aprecords.htm#minimal, and $p+2d$ is just a beginning of an AP-23.
Now, one sees immediately that $p, p+6d, p+10d$ form a simplex, as it is a subsequence of the arithmetic progression of primes $p, p+2d, p+4d, p+6d, p+10d$.
Similarly, $p, p+6d, p+15d$ form a simplex, as do $p, p+10d, p+15d$ and $p+6d, p+10d, p+15d$. These come from sequences with differences $3d$, $5d$, and $d$ respectively.
However, $p,p+6d,p+10d,p+15d$ can only form a simplex if they all lie in some arithmetic progression of primes. But then its difference must divide $d$, contradicting the fact that $p+d$ is not prime.