Solving recurrence relation
The cost at each level isn't $npk$. Let's look at the first few levels.


The cost at the top level, $l=0$, is indeed $npk$.
At the next level ($l=1$) you have three recursive calls: $T(n,p,k/2)$, $T(n,p/4,k)$, and $T(n/8,p,k)$ with costs $npk/2, npk/4, npk/8$ for a total cost at level two of
$$
\frac{npk}{2}+\frac{npk}{4}+\frac{npk}{8}=npk\cdot\frac{7}{8}
$$
At the subsequent level you have nine recursive calls and if you do a bit of arithmetic you see that the total cost at that level ($l=2$) is
$$
npk\cdot\left(\frac{7}{8}\right)^2
$$ 
You can show inductively that the cost at level $l$ will be
$$
npk\cdot\left(\frac{7}{8}\right)^l
$$
so the total cost up to and including level $M$ will be
$$
T(n,p,k)=\sum_{l=0}^Mnpk\cdot\left(\frac{7}{8}\right)^l=npk\cdot\frac{1-(7/8)^{M+1}}{1-7/8}&lt;8npk
$$
since, as $M$ increases $(7/8)^{M+1}\rightarrow 0$. In other words, $T(n,p,k)=O(npk)$


It's clear that $T(n,p,k)&gt;npk$, since that's the contribution at level zero, so we have, finally, $T(n,p,k)=\Theta(npk)$.