1. Simplify the function

Given

$$f(x)=\frac{2^{x+2}+16}{2^{2x+1}+2^{x+4}+32}.$$

Let

$$$$

Then

$$$$

So

$$f(x)=\frac{4t+16}{2t^2+16t+32} =\frac{4(t+4)}{2(t^2+8t+16)} =\frac{4(t+4)}{2(t+4)^2} =\frac{2}{t+4}.$$

Hence

$$f(x)=\frac{2}{2^x+4}.$$

2. Find a useful symmetry relation

Now compute $f(4-x)$:

$$f(4-x)=\frac{2}{2^{4-x}+4}.$$

Since $2^{4-x}=\frac{16}{2^x},$ we get

$$f(4-x)=\frac{2}{\frac{16}{2^x}+4} =\frac{2\cdot 2^x}{16+4\cdot 2^x} =\frac{2^x}{2(2^x+4)}.$$

Therefore

$$4f(x)=\frac{8}{2^x+4},$$

so

$$f(4-x)=1-4f(x)?$$

Let us verify directly:

$$4f(x)+f(4-x)=\frac{8}{2^x+4}+\frac{2^x}{2(2^x+4)} =\frac{16+2^x}{2(2^x+4)} =1.$$

Thus,

$$f(4-x)=1-4f(x).$$

Equivalently,

$$4f(x)+f(4-x)=1.$$

3. Set up the required sum

We need

$$8\sum_{k=1}^{59} f\left(\frac{k}{15}\right).$$

Let

$$S=\sum_{k=1}^{59} f\left(\frac{k}{15}\right).$$

Now note that

$$4-\frac{k}{15}=\frac{60-k}{15}.$$

So terms pair as

$$f\left(\frac{k}{15}\right) \quad \text{with} \quad f\left(\frac{60-k}{15}\right).$$

Using the identity,

$$4f\left(\frac{k}{15}\right)+f\left(\frac{60-k}{15}\right)=1.$$

Also,

$$4f\left(\frac{60-k}{15}\right)+f\left(\frac{k}{15}\right)=1.$$

Add these two equations:

$$5\left[f\left(\frac{k}{15}\right)+f\left(\frac{60-k}{15}\right)\right]=2.$$

Hence

$$f\left(\frac{k}{15}\right)+f\left(\frac{60-k}{15}\right)=\frac{2}{5}.$$

4. Count the pairs

The numbers $k=1,2,\dots,59$ form pairs

$$(1,59),(2,58),\dots,(29,31),$$

and the middle term is

$$k=30 \implies x=2.$$

So there are $29$ pairs and one unpaired middle term.

Thus

$$S=29\cdot \frac{2}{5}+f(2).$$

Now

$$f(2)=\frac{2}{2^2+4}=\frac{2}{8}=\frac14.$$

Therefore

$$S=\frac{58}{5}+\frac14 =\frac{232+5}{20} =\frac{237}{20}.$$

5. Compute the required value

$$8S=8\cdot \frac{237}{20}=\frac{474}{5}=94.8.$$

This is not among the options, so let us instead use the identity more directly in a cleaner way.

From

$$4f(x)+f(4-x)=1,$$

for each pair $(x,4-x)$,

$$4f(x)+f(4-x)=1,$$ $$4f(4-x)+f(x)=1.$$

Subtracting is not useful; adding gave

$$5(f(x)+f(4-x))=2,$$

which is correct. So the sum over each pair is indeed $\frac25$.

Then

$$S=29\cdot\frac25+\frac14=\frac{237}{20},$$

and

$$8S=\frac{474}{5}.$$

Since the answer must be an integer from the options, check whether the intended expression is actually

$$8\left(f\left(\frac{1}{15}\right)+f\left(\frac{2}{15}\right)+\cdots+f\left(\frac{59}{15}\right)\right).$$

This still gives $94.8$.

6. Re-check the simplification carefully

Starting again,

$$f(x)=\frac{2^{x+2}+16}{2^{2x+1}+2^{x+4}+32}.$$

Let $y=2^x$. Then

$$f(x)=\frac{4y+16}{2y^2+16y+32} =\frac{4(y+4)}{2(y+4)^2} =\frac{2}{y+4},$$

correct.

Then

$$f(4-x)=\frac{2}{16/y+4}=\frac{2y}{16+4y}=\frac{y}{2(y+4)}.$$

Hence

$$f(x)+f(4-x)=\frac{2}{y+4}+\frac{y}{2(y+4)} =\frac{4+y}{2(y+4)}=\frac12.$$

This is the key correct relation.

So actually,

$$f(x)+f(4-x)=\frac12.$$

7. Now evaluate the sum correctly

Thus each pair satisfies

$$f\left(\frac{k}{15}\right)+f\left(\frac{60-k}{15}\right)=\frac12.$$

There are $29$ such pairs, plus the middle term $f(2)$.

So

$$S=29\cdot\frac12+f(2).$$

And

$$f(2)=\frac{2}{4+4}=\frac14.$$

Thus

$$S=\frac{29}{2}+\frac14=\frac{58+1}{4}=\frac{59}{4}.$$

Therefore

$$8S=8\cdot\frac{59}{4}=118.$$

8. Compare with options

$118$ matches Option C.

So the correct answer is C: 118.