$f$
is defined recursively for all integers $n>1,p>0$
as follows: $$ \begin{aligned} f(1,p) &= C + p, \\ f(n,1) &= C, \\ f(n,p) &= f(n1, f(n,p1)) \end{aligned} $$
where $C>0$
is a real constant. Give a nonrecursive expression for $f(4,p).$
Proof is not required.
Hint: You may want to try $f(2,p)$
first.

Try writing out and spotting patterns in
$f(2,p).$

Try writing this in a nonrecursive form.

… by noticing it is only recursive in the second argument,
$p$
(the$2$
is fixed). 
How might you do the same for
$f(3,p)?$

Remember to use results you’ve already shown.

Using the hint provided in the question, we try to find an expression for
$f(2,p).$
From the definition we get that$f(2,p) = f(1, f(2, p1)) = C + f(2, p1).$
Since the first argument is fixed at$2,$
we can unravel it to get$f(2,p)=C+\ldots+C+f(2,1)=C \cdot p.$
We can apply the same strategy to find
$f(3,p).$
We start with$f(3,p) = f(2, f(3, p1)),$
at which point we use our previous result to get$f(3,p)=C \cdot f(3,p1).$
Similarly, this unravels to$f(3,p)=C^p.$
Finally, we get to
$f(4,p) = f(3, f(4, p1)) = C^{f(4,p1)},$
which expands to$f(4, p) = C^{C^{\cdot^{\cdot^{\cdot^C}}}},$
which is a tower of height$p.$