F[A,B,C,D]=πm(0,2,5,7,8,10,12,15),the expression is in POS i.e Product of Sum.

Quine-McCluskey method is as follows:

Step 1:

Write the binary equivalent of the numbers given in the problem{For example 1 is written as 0001,10 is written as 1010}.Among the obtained binary numbers group the ones having no 1's,1 1's,2 1's,3 1's and 4 1's.{That forms the first column in the given table}.

Step 2:

Different groups formed in the first column have to be compared with one another.The first group should be compared with the second group,second group should be compared with the third and so on.While doing so, replace 0&0 with a 0,1&1 with a 1 and 0&1 with a hypen(-) or dash.In the second column write the ones which on comparing give only one dash {For example 00-1}

Step 3:

The obtained groups in the second column are again compared with one another.While doing so, replace 0&0 with a 0,1&1 with a 1 and 0&1 with a hypen(-) or dash.In the third column write the ones which on comparing give only two dash {For example -0-1}.Repeat the steps until no increment of dashes is possible.

The initial table can be shown as follows

10-0(8,10) 1-00(8,12) | | | 0101(5) 1010(10) 1100(12) | 01-1(5,7) | | | 0111(7) | -111(7,15) | | | 1111(15) | | |  

Since no other comparisions can be made, consider the values at the end of third column.The 1's are wriiten as a compliment of an alphabet for example $\overline A$ and 0's are written as normal alphabet for example A and -'s are just ignored.If the same numbers are repeated then just consider them once.Since its the product of sums, each group alphabets are written as sum and all the sums of groups are written as a product of one another.Consider those and check wether any of the numbers in a group can be replaced by any other group,if yes ignore that group.

This can be shown as follows

In this there is only one group so retain it and that forms the minimal expression.

Therefore f(A,B,C,D)=${B}+D$