Pass shell variable as a /pattern/ to awk

Use awk's ~ operator, and you don't need to provide a literal regex on the right-hand side:

function _process () {
    awk -v l="$line" -v pattern="$1" '
        $0 ~ pattern {p=1} 
        END {if(p) print l >> "outfile.txt"}
    '  
}

Although this would be more efficient (don't have to read the whole file)

function _process () {
    grep -q "$1" && echo "$line"
}

Depending on the pattern, may want grep -Eq "$1"

awk  -v pattern="$1" '$0 ~ pattern'

Has an issue in that awk expands the ANSI C escape sequences (like \n for newline, \f for form feed, \\ for backslash and so on) in $1. So it becomes an issue if $1 contains backslash characters which is common in regular expressions (with GNU awk 4.2 or above, values that start with @/ and end in /, are also a problem). Another approach that doesn't suffer from that issue is to write it:

PATTERN=$1 awk '$0 ~ ENVIRON["PATTERN"]'

How bad it's going to be will depend on the awk implementation.

$ nawk -v 'a=\.' 'BEGIN {print a}'
.
$ mawk -v 'a=\.' 'BEGIN {print a}'
\.
$ gawk -v 'a=\.' 'BEGIN {print a}'
gawk: warning: escape sequence `\.' treated as plain `.'
.
$ gawk5.0.1 -v 'a=@/foo/' BEGIN {print a}'
foo

All awks work the same for valid escape sequences though:

$ a='\\-\b' awk 'BEGIN {print ENVIRON["a"]}' | od -tc
0000000   \   \   -   \   b  \n
0000006

(content of $a passed as-is)

$ awk -v a='\\-\b' 'BEGIN {print a}' | od -tc
0000000   \   -  \b  \n
0000004

(\\ changed to \ and \b changed to a backspace character).

Try something like:

awk -v l="$line" -v search="$pattern" 'BEGIN {p=0}; { if ( match( $0, search )) {p=1}}; END{ if(p) print l >> "outfile.txt" }'