Leaving out the $S's$ for the moment, there are $\frac{7!}{4!2!} = 105$ permutations of $PPMIIII$
Of these $\frac{6!}{4!} = 30$ will have the $P's$ together, thus $75$ will have the $P's$ apart.
For the together cases, we need to borrow an $S$ to separate them, e.g. $\boxed{PSP}MIIII$
The remaining $3\; S's$ can be inserted in the gaps between units (including ends) in $\binom73$ ways
For cases with $P's$ already separate, e.g. PMPIIII, the $4\; S's$ can be inserted similarly in $\binom84$ ways.
Thus permissible permutations $= 30\binom73 + 75\binom84 = 6300$
For better or worse, this give a value of $k=1$
Use of heavy artillery
To make assurance doubly sure, I have used Jair Taylor's formula to confirm the answer.
Define polynomials for $k\geq 1$ by $q_k(x) =
\sum_{i=1}^k \frac{(-1)^{i-k}}{i!} {k-1 \choose i-1}x^i$.
e.g. for $k=2, q_2(x)$ works out to ${(x^2-2x)}/2!$
For the nonce, we shall take the $I's$ to be distinct, divide by $4!$ at the end of the calculations. So we have five distinct alphabets, one of length $2$ and one of length $4$
The number of permutations will be given by
$$\int_0^\infty \prod_j q_{k_j}(x)\, e^{-x}\,dx.$$
Wolframalpha gives the answer as $151,200$
Dividing by $4!$ to take care of the $4 I's$, we again get the answer of $6300$
It's an application of the inclusion exclusion principle. We use the above to write properties $P_1, P_2$ and $P_3$ respectively as
$P_1$ - Pattern CAN occurs
$P_2$ - Pattern BIN occurs
$P_3$ - Pattern NIB occurs
Let $A_1, A_2, A_3$ be the sets conforming to each property. Now using the above stated principle we know that
$$|\bar{A_1}\cap\bar{A_2}\cap\bar{A_3}| = |S| - \sum_{i=1}^3|A_i| +\sum_{i,j}|A_i \cap A_j| - |A_1\cap A_2\cap A_3|$$
$A_1$ - Treat CAN as one entity and calculate the permutations of the entities (CAN),O,M,B,I,T,I,O,N which is equal to
$$|A_1| = \frac{9!}{2!}$$
$A_2$ - Treat BIN as one entity and calculate the permutations of the entities C,O,M,(BIN),A,T,I,O,N which is equal to
$$|A_2| = \frac{9!}{2!}$$
$A_3$ - Treat NIB as one entity and calculate the permutations of the entities C,O,M,(NIB),A,T,I,O,N which is equal to
$$|A_3| = \frac{9!}{2!}$$
$A_1\cap A_2$ - Putting CAN and BIN together and calculating the permutations of the entities (CAN),(BIN),O,M,I,O,T which is equal
$$|A_1\cap A_2| = \frac{7!}{2!}$$
$A_2\cap A_3$ - We cannot have BIN and NIB together as there's only one B in COMBINATION
$$|A_1\cap A_3| = 0$$
So, also $|A_1\cap A_2\cap A_3| = 0$
EDIT - As was suggested by the commenter, there's one extra case that I forgot to account for, which is CANIB. This will be a part of $A_1\cap A_3$ and is calculated by the number of permutations of (CANIB),O,O,T,I,N,M
$$|A_1\cap A_3| = \frac{7!}{2!} + \frac{7!}{2!}$$
Hence the total number of combinations become
$$|\bar{A_1}\cap\bar{A_2}\cap\bar{A_3}| = \frac{11!}{2!2!2!} - 3\cdot\frac{9!}{2!} + 3\cdot\frac{7!}{2!}$$
Best Answer
If you have $\_B\_N\_N\_S\_$ you can allocate three $A$'s among any of those $5$ empty spaces. That's $\binom{5}{3} = 10$ ways to allocate the $A$'s.
You then multiply that by the number of ways to arrange $B, N, N, S$ amongst themselves, which you have already done: $\frac{4!}{2!} = 12$.
All in all, that's $\binom{5}{3} \cdot \frac{4!}{2!} = 10 \cdot 12 = 120$ ways to arrange $BANANAS$ with all the $A$'s separated.