is the set of integers as big as P(prime numbers)? I have a function $f:\mathbb{N}\to P(\mathbb{P})$ where $f(x) = \{prime\ factors\ of\ x\}$

$f$ is not $1:1$ (easy)

I proved that $f$ is on $P(\mathbb{P})$ which is probably wrong since it means $|\mathbb{N}|\geq |P(\mathbb{P})|$ but $|\mathbb{N}|=|\mathbb{P}|$.

Every subset of primes has a source in $\mathbb{N}$ (the multiplication of all the primes)

What's wrong with this?
Every finite set of primes is in the range of $f$.

However, what about any infinite set of primes? Say, the set of all primes, itself?

No number has infinitely many prime factors, so no infinite set of primes is in the range of $f$. Indeed, the set of finite sets of primes is countable, while the set of infinite sets of primes is uncountable - "most" sets of primes are infinite!
As someone pointed out in the comments, $\mathbb{N}$ has the same cardinality as $\mathbb{P}$ because they are both infinite subsets of the countably infinite set $\mathbb{Z}$, hence they are both countably infinite sets. If you want to see this explicitly, just define $f : \mathbb{N} \to \mathbb{P}$ letting $f(n)$ be the $n$th prime, i.e. $f(1) = 2, f(2) = 3, f(3) = 5,\ldots$.