The following function should search a character inside string.

A1) & takes the address of the object to its right, so &s[0] takes the address of s[0], which is identical to the value of s

A2) pc[i] is the value of the ith element of the array pc, not the value of address of ith element. In this case, given that pc is a pointer to a char, then pc[i] is equivalent to *(pc + i), so *pc[i] is the equivalent to *(*(pc+i)). Since the value of pc[i] can only be in -128...127, the value of *pc[i] is likely to cause a program crash, since, except for the case where pc[i] == 0, it is most likely that you do not "own" the address in question.

A3) See A2: in short pc[i] is a character, not an address

The code given suggest that you do not understand pointers very well. Most C programmers would have written this something like

char *find_char(char c, const char *s)
{
     while(*s) {
       if( *s == c )
           return s;
       s += 1;
     }
     return NULL;
}

Note that we declare s as const char *This does 2 things, 1) the complier will make sure that we don't accidentally assign a value to s, and 2) since the compiler knows that the values that s points to can't be changed, it may do some optimizations that it would not be able to do otherwise. Also, const char *s declares a pointer to const char, which means that we can move s, but not assign a value to it. If we wanted to declare a pointer that could not be moved, but we wanted to be able to change the value that it points to, that would be char * const s. Logically then, const char * const s is a const pointer to const char, i.e. we can't move the pointer, or assign a new value to what it points to.
Since we do have a pointer to a string of characters that ends in'\0', we can simply move the pointer forward until we either find the char we are interested in *s == c or we get to the end of the string *s == '\0'. I could have written while(*s != '\0') in the while loop, but I've taken advantage of the fact that in C/C++ any non-zero value is treated as true. So if the passed in array is "Hello World!", then the value of the s[0] is 'H', decimal 72, which is not zero, and so on until after the '!', when the ending character is '\0' which has the decimal value of zero.
Actually, though an experienced C programmer would probably call strchr(), which would might make use of assembler instructions that are optimized for this type of thing.