I have given the Weierstraß p-adic preparation theorem. It is stated as follows: Let $f=a_0 + a_1T + ... \in \mathbb{Z}_p[[T]]$ for a prime $p$ such that $p \mid a_0,...,a_{n-1}$ and $p\not\mid a_n$. Then there exists a unique unit $U\in\mathbb{Z}_p[[T]]^*$ and a distinguished polynomial P of degree $n$, i.e. $P= b_0 + b_1 T + ... + b_{n-1}T^{n-1}+T^n$ with $p\mid b_0,...,b_{n-1}$ such that $f=PU$.
The existence is clear to me but the uniqueness not. Both supposedly follow from a prior theorem, division with rest. It is stated as follows: Let $f=a_0 + a_1T + ... \in \mathbb{Z}_p[[T]]$ for a prime $p$ such that $p \mid a_0,...,a_{n-1}$ and $p\not\mid a_n$. Then, for all $g\in\mathbb{Z}_p[[T]]$ there is a unique expression $$ g = qf + r $$ with $ q \in \mathbb{Z}_p[[T]]$ and $ r $ a polynomial of degree $ \deg(r) \le n-1 $.
The uniqueness statement that I require is a rather strong one. I need that if $f=PU$ as in the first theorem and also $f=P'U'$ with $U'$ a unit in $\mathbb{Z}_p[[T]]$ and $P'$ a polynomial, then $P=P'$ and $U=U'$. My problem is that P' might have a different degree than $P$, in which case I do not see how to apply division with rest.
I need this strong uniquement because I would like to conclude from this theorem that $\mathbb{Z}_p[[T]]$ is a UFD.
Help would be greatly appreciated!
As @BadamBaplan says, your desired uniqueness does not hold. Consider, for instance, the polynomial $f=p+X^n+pX^{n+1}$. Then we have the factorization $U'=1$, $P'=f$, and the Weierstraß factorization $f=UP$, where indeed, $P$ and $U$ are given to you already by Hensel’s Lemma. If I understand your question aright, the factorization $f=U'P'$ satisfies your criteria.