1st point:Though I am not php expert, I have seen usually shell commands run under exec command.
2nd point: You need not to use cat with awk. awk could read any Input_file passed to it. eg--> awk '........' Input_file
A try to fix your code: Since you have NOT mentioned why you are printing x variable again in awk program + why date is mentioned in variable but not being used?
What I am doing here is calling php script by passing $9 with escaping $ in it by awk's system command(Without samples I am not able to run this command).
awk -v s1="'" -v s2="\"" -v date="$(date +"%Y-%m-%d %H:%M:%S")" '
{
system(s1 "/usr/bin/php-cgi -f /path/hex2bin.php param1= " s2 "$9" s1)
}' Input_file
But 1 things is there using system command in awk you will be able to execute other/shell commands by it, since it is executing commands in another sub shell so you will NOT be able to take its output value into awk variable, experts could correct me if I missed something here.
EDIT: Or try to use while loop itself.
while read line one two three four five six seven eight nine ten
do
new_variable=$(your_php_command "$nine")
echo "Use here new_variable...."
done < "Input_file"
cat file |ibm.com/developerworks/aix/library/au-badunixhabits.html#ten