how to pass a parameter to char*?

This literal "TESTEEEE" is of type char const[9]. When used as an argument to a function, it can decay to char const* but not to char*. Hence to use your function, you have to make the parameter fit to your argument or the opposite as follows

#include <iostream>

using namespace std;
int PrintString(const char* s)
{
    cout << s << endl;
}


int main(int argc, char* argv[])
{

    PrintString("TESTEEEE");

    return 0;
}

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OR

#include <iostream>

using namespace std;
int PrintString( char* s)
{
    cout << s << endl;
}


int main(int argc, char* argv[])
{
    char myArr[] = "TESTEEEE";
    PrintString(myArr);

    return 0;
}

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