Extract column using grep

First figure out the command to find the column number.

columnname=C
sed -n "1 s/${columnname}.*//p" datafile | sed 's/[^\t*]//g' | wc -c

Once you know the number, use cut

cut -f1,3 < datafile 

Combine into one command

cut -f1,$(sed -n "1 s/${columnname}.*//p" datafile | 
   sed 's/[^\t*]//g' | wc -c) < datafile

Finished? No, you should improve the first sed command when one header can be a substring of another header: include tabs in your match and put the tabs back in the replacement string.

You need to use awk:

awk '{print $1,$3}' <namefile>

This simple command allows printing the first ($1) and third ($3) column of the file. The software awk is actually much more powerful. I think you should have a look at the man page of awk.

A nice combo is using grep and awk with a pipe. The following code will print column 1 and 3 of only the lines of your file that contain 'p1':

grep 'p1' <namefile> | awk '{print $1,$3}'

If, instead, you want to select lines by line number you can replace grep with sed:

sed 1p <namefile> | awk '{print $1,$3}'

Actually, awk can be used alone in all the examples:

awk '/p1/{print $1,$3}' <namefile> # will print only lines containing p1
awk '{if(NR == 1){print $1,$3}}' <namefile> # Will print only first line