C++ Strings – How to Convert Single Char to Null-Terminated const char*

There are a few things I don't understand here. I see that this is tagged for C++/CLI, but what I describe below should be the same as Standard C++.

Doesn't compile

The code you give doesn't compile; get_error_from_header does not specify a return type. In my experiments I made the return type size_t.

Signature of C++ strstr()

The signature for strstr() in the standard C library is:

char *
strstr(const char *s1, const char *s2);

but the signature for strstr() in the C++ library, depending on the overload, is one of:

const char * strstr ( const char * str1, const char * str2 );
      char * strstr (       char * str1, const char * str2 );

I would choose the first overload, because you don't want to modify the string, you only want to read it. Therefore you can change your code to:

const char* err = strstr((const char *)ptr, "550");
if (err != NULL) {
    ...
}

Also, I'm assuming your comment reporting the error:

//error  cannot convert const char** to char*

is a typo: there's no const char** to be seen.

Assignment to err unnecessary

As is pointed out in a previous answer, the use of err to store the result of strstr is unnecessary if all it's used for is checking NULL. Therefore you could use:

if (strstr((const char *)ptr, "550") != NULL) {
    ...
}

Use of reinterpret_cast<> encouraged

As is pointed out in another answer you should be using reinterpret_cast<> instead of C-style casts:

if (strstr(reinterpret_cast<const char *>(ptr), "550") != NULL) {
    ...
}

Use of const_cast<> to strip const

Given the example in the question, I don't see where this is necessary, but if you had a variable that you need to strip of const-ness, you should use the const_cast<> operator. As in:

const char * p1;
char * p2;

p2 = const_cast<char *>(p1);

As is pointed out in a comment, the reason to use const_cast<> operator is so that the author's intention is clear, and also to make it easy to search for the use of const_cast<>; usually stripping const is the source of bugs or a design flaw.

printMe takes an lvalue reference to a mutable pointer to const char.

In your first example, tmp is an lvalue of type mutable pointer to const char, so a reference can be bound to it without issue.
In your second example, (const char*)s creates a temporary const char* object. Lvalue references to mutable objects can't bind to temporaries, so you get an error. If you change printMe to take a const char* const& then the call will succeed with or without the explicit cast.

void printMe(const char * const& buf) {
    printf("Given Str = %s", buf);
}

int main() {
    char s[] = "test string";
    printMe(s);
}

Live on Coliru

Of course, if you don't want to alter the object (the pointer) passed into printMe, then there's no reason to use a reference at all. Just make it take a const char*:

void printMe(const char * buf) {
    printf("Given Str = %s", buf);
}

int main() {
    char s[] = "test string";
    printMe(s);
}

Live on Coliru

In the end, this is the same reason something like this:

void doSomething(const std::string& s) {}
int main() {
    doSomething("asdf");
}

works while this:

void doSomething(std::string& s) {}
int main() {
    doSomething("asdf");
}

does not. A temporary object is created, and the reference to non-const object can't bind to the temporary.