$P(4,1)$  is true because of $2,3,4,5$ but I strongly doubt that $P(5,1)$ is true. Indeed $P(4,1)$ is only true that once and even $P(3,1)$ happens for the last time at $7,8,9$. 
Of any $210$ consecutive integers one is divisible by $2,3,5,7$ so $P(210,3)$ fails. I'm sure that can be improved. It does generalize .

As noted in the comments, there are two cases (at least) of $8$ consecutive integers each with two distinct prime factors and a run of $12$ can be ruled out. Actually it seems extremely unlikely that there are any cases of $9$, let alone $10$ or $11$: Such a run of $9$ would have to include $7$ integers $3(a-1),2(b-1),*,2^i3^j,*,2(b+1),3(a+1)$ where  $a\pm1=2^{i}3^{j-1}\pm1$ and $b\pm1=2^{i-1}3^{j}\pm1$ are two pairs of twin primes (or merely twin prime powers). I would expect only finitely many cases where $2^u3^v\pm1$ are both prime powers (I find $35$ such up to $10^{50}$ with $u,v  \ge 1$, let alone getting two such pairs as above ( I find $4$ cases, $36,144,216$ and $2^{33}3^9$.) This leaves out the requirements that we would also need $2^{i}3^{j}\pm1$ to each have two distinct prime factors (ruling out the smallest and largest) and have at least one of $2^{i}3^{j}\pm4$ to be 4 times a prime power, and more.