I'm not sure I get it right.
If our task is to give a distinct one-digit numerical value to all the letters so that all the equations hold, then this might work (although not the only solution - there are 2x2x3!x3!x3x3!=2592 of them, I won't post them all):
a=9; c=8; e=2; i=4; k=3; l=7; o=1; p=0; s=5; t=6
Then
pepsi $=0\times2\times0\times5\times4=0$
cocacola $=8\times1\times8\times9\times8\times1\times7\times9=290304$
cocktail $=8\times1\times8\times3\times6\times9\times4\times7=290304$
So, no matter how many times you add pepsi (0) to the cocacola (290304), it still remains cocktail (290304).
However, if my interpretation of products is wrong, and @Marius has the good approach with concatenation, then the possible answers are these:
n pepsi cocacola cocktail
1 42497 6010681 6053178
1 24296 8010871 8035167
1 26294 8010851 8037145
1 47492 6010631 6058123
1 40497 26212681 26253178
1 43489 27212701 27256190
1 42489 37313701 37356190
1 20296 48414871 48435167
1 20294 68616851 68637145
1 40492 76717631 76758123
1 75781 4020432 4096213
1 34378 19121902 19156280
1 31378 49424902 49456280
1 70781 54525432 54596213
1 19183 70727052 70746235
1 17183 90929052 90946235
1 51563 8040874 8092437
1 50563 18141874 18192437
1 15123 46474607 46489730
1 14123 56575607 56589730
2 15192 76707640 76738024
2 4083 69616971 69625137
2 6089 52535213 52547391
2 27264 58535813 58590341
2 6097 12141284 12153478
2 8021 34353475 34369517
3 4078 39313951 39326185
3 20239 54515481 54576198
3 27204 65616531 65698143
3 26204 85818531 85897143
3 12176 80858035 80894563
4 18153 64606420 64679032
4 10179 35323582 35364298
4 10173 85828542 85869234
4 8014 52535293 52567349
4 10137 59545924 59586472
4 7069 43454315 43482591
4 1028 94959475 94963587
4 7013 45464586 45492638
4 3071 45484528 45496812
5 12149 73707350 73768095
5 13169 82808250 82874095
5 7015 42434283 42469358
5 1075 69636983 69642358
Where n stands for which equation we are referring to: n pepsi + 1 cocacola = 1 cocktail.
I posted answers which have leading zeros, if those are considered invalid, just neglect them.
I did not come up with these manually, used a few lines of python code to get them:
import itertools as it
for n in range(1,6):
for (a,c,e,i,k,l,o,p,s,t) in it.permutations(range(10)):
pepsi = 10100*p+1000*e+10*s+i
cocacola = 10101000*c+1000100*o+10001*a+10*l
cocktail = 10100000*c+1000000*o+10000*k+1000*t+100*a+10*i+l
if n*pepsi+cocacola==cocktail:
print '{0:1d} {1:5d} {2:8d} {3:8d}'.format(n, pepsi, cocacola, cocktail)