Destroying the P-filter-property It is known that if a forcing notion is proper, then every P-filter will generate a P-filter in the generic extension (see, e.g., Shelah, Proper and Improper Forcing, VI.5)
On the other hand, if we start collapsing cardinals, we can destroy the P-filter property. For example, making a base of a P-point countable will add a sequence of elements from the filter (namely, a complete enumeration of that base) that serves as a counterexample for the P-filter property in the extension.
So my question is:

Are there examples of "nicer" (e.g., not collapsing cardinals) forcing notions that destroy P-filters in this way, i.e., add a sequence in the filter that the filter cannot decide?
More spectacularly, is there maybe a forcing notion that could preserve a P-point as an ultrafilter while destroying the P-filter property?

EDIT: As Martin Goldstern pointed out, I should add that I'm interested in filters on $\omega$.
 A: (A very partial answer.) Assuming CH, the answer seems to be "no" (for filters on countable sets).
CH implies that every P-filter on $\omega$ is generated by a tower (i.e., an almost decreasing sequence) of length $\omega_1$.  In any $\omega_1$-preserving forcing extension any countable sequence of filter sets can be refined to a countable sequence of sets from the tower, so there is a lower bound in the tower. 
A: A second partial (and not very deep, sorry) answer:  Assume that at least one of the following holds: 


*

*zero sharp does not exist, i.e., Jensen's covering lemma holds. 

*The continuum is below $\aleph_\omega$ (or at least: your filter is generated by less than $\aleph_\omega$ many sets). 
Then the answer is again "no" (if the forcing is not allowed to collapse cardinals). 
Proof:  Every new countable subset of the filter base is contained in an old countable subset of the filter base. (This is well known, but for the sake of completeness I give a sketch of the proof.   Fix a bijection between the filter base and some ordinal $\alpha$.  Every new countable set $A \subseteq \alpha$  is contained in an old set $B$ of size $\aleph_n$, for some $n$. [Under assumption 1, the covering lemma gives $n\le 1$; assumption 2 just says outright that there is some $n$.]   Now fix $A$ and  choose $n$ as small as possible.  Using a bijection from $B$ to $\aleph_n$ in $V$, we may wlog assume that the original set $A$ was a subset of $\omega_n$.   If $n>0$, then the countable set  $A$ is bounded in $\omega_n$, so we can cover $A$ by an ordinal of cardinality $\aleph_{n-1}$, contradiction.)
