why does a const char don't need a memory address to point to?*

It does.

A C-string literal like

"Anthony"

is decayed to the address of its 1^st character. Like, BTW; any array in C does.

You can imagine this declaration

const char* a = "Anthony";

the following way

const char string_literal[] = "Anthony";

const char *a = string_literal;

That is the compiler creates an array of characters with the static storage duration that stores the string "Anthony" and the address of the first character of the array (due to the implicit conversion of array designators to pointers to their first characters) is assigned to the pointer a.

Here is a demonstrative program that shows that string literals are character arrays.

#include <iostream>
#include <type_traits>

decltype( auto ) f()
{
    return ( "Anthony" );
}

template <size_t N>
void g( const char ( &s )[N] )
{
    std::cout << s << '\n';
}

int main() 
{
    decltype( auto ) r = f();

    std::cout << "The size of the referenced array is "
              << std::extent<std::remove_reference<decltype( r )>::type>::value
              << '\n';

    g( r );

    return 0;
}

The program output is

The size of the referenced array is 8
Anthony

The size of the string literal (of the array that stores the string literal) is equal to 8 because the string includes also the terminating zero character '\0'.

In the demonstrative program the expression

std::extent<std::remove_reference<decltype( r )>::type>::value

may be substituted for just the expression

sizeof( r )

"Why does const char* don't need a pointer to a memory address?"

In fact, It does need an memory address to point to.

const char* a means a is a pointer to a string literal or character constant.

A pointer always requires an address to point to because it is the nature of a pointer to point to a specific object in memory. So, a and any other pointer to const char does too.

A string literal like "Hi My Name is Alfred!" by an assignment like:

const char* a;
a = "Hi My Name is Alfred!";

decays to a pointer to the address of the first element of the string literal.

Means in turn, a gets assigned by the address of the first element of the string literal "Hi My Name is Alfred!" which could be stored anywhere in memory dependent on the execution environment.

It is not in the might of a programmer where a string literal is exactly stored. Your assignment is only to assign and handle the respective pointer appropriately.

It does need a memory address, and it HAS a memory address. In your example it's simply the memory address of the beginning of the string. It's the same with any other array variable that's initialized at compile time, for instance "int array [] = {0, 1, 2, 3};".

If you used a binary editor to look at the executable, you would see the string "Anthony" in there. If you put the line "printf ("a is at %p\n", (void *)a);" in your program, then compile & run it, you'll see the address.