In the corrected problem there are still $16!$ sequences altogether. The number that include the letters PONK in that order (but not necessarily adjacent) can be computed as follows: there are $\binom{16}4$ ways to choose the four positions to be filled by the letters P, O, N, and K, and there are then $12!$ ways to permute the remaining letters, so there are $\binom{16}412!$ sequences that contain ...P...O...N...K... .
In similar fashion you can count the sequences that contain ...D...O...B...A... and those that contain ...C...O...P... . However, the sets $\{\text{P,N,K}\}$, $\{\text{D,B,A}\}$, and $\{\text{C,P}\}$ are pairwise disjoint, so a sequence can contain two or even all three of these subsequences, and we’ll need to go through the complete inclusion-exclusion calculation.
Let’s try to count the sequences that contain both ...P...O...N...K... and ...D...O...B...A... . First we choose $7$ of the $16$ places for the letters P, O, N, K, D, B, and A; this can be done in $\binom{16}7$ ways. The O must fill the third of these seven positions. The P and D must fill the first two, but they can do so in either order. There are $\binom42$ ways to choose the two positions for N and K, which must go in that order, and B and A will then fill the remaining positions, again in that order. Finally, the other $9$ letters can be permuted arbitrarily. The total is therefore
$\binom{16}7\cdot2\cdot\binom42\cdot9!\;.$
You should be able to use similar analyses to calculate the remaining terms of the inclusion-exclusion calculation.