The issue is that you are not allocating any space for those names. You need to initialize each element in the array if you intend to use it with scanf.

char* names[6];
for( int i = 0; i < 6; ++i )
    names[i] = malloc( 256 * sizeof *names[i] ); // or some other max value

scanf( "%s", names[1] );

Otherwise those pointers will be pointing anywhere in your memory, and attempting to read/write those locations will eventually result in a segmentation fault.

There are two way of working with array of characters (strings) in C. They are as follows:

char a[ROW][COL];
char *b[ROW];

Pictorial representation is available as an inline comment in the code.

Based on how you want to represent the array of characters (strings), you can define pointer to that as follows

    char (*ptr1)[COL] = a;
    char **ptr2 = b;

They are fundamentally different types (in a subtle way) and so the pointers to them is also slightly different.

The following example demonstrates the different ways of working with strings in C and I hope it helps you in better understanding of array of characters (strings) in C.

#include <stdio.h>
#include <string.h>
#include <stdlib.h>

#define ROW 5
#define COL 10

int main(void) 
{
    int i, j;
    char a[ROW][COL] = {"string1", "string2", "string3", "string4", "string5"};
    char *b[ROW];

    /*

    a[][]

      0   1   2   3   4   5   6     7    8   9
    +---+---+---+---+---+---+---+------+---+---+
    | s | t | r | i | n | g | 1 | '\0' |   |   |
    +---+---+---+---+---+---+---+------+---+---+
    | s | t | r | i | n | g | 2 | '\0' |   |   |
    +---+---+---+---+---+---+---+------+---+---+
    | s | t | r | i | n | g | 3 | '\0' |   |   |
    +---+---+---+---+---+---+---+------+---+---+
    | s | t | r | i | n | g | 4 | '\0' |   |   |
    +---+---+---+---+---+---+---+------+---+---+
    | s | t | r | i | n | g | 5 | '\0' |   |   |
    +---+---+---+---+---+---+---+------+---+---+

    */  

    /* Now, lets work on b */    
    for (i=0 ; i<5; i++) {
        if ((b[i] = malloc(sizeof(char) * COL)) == NULL) {
            printf("unable to allocate memory \n");
            return -1;
        }
    }

    strcpy(b[0], "string1");
    strcpy(b[1], "string2");
    strcpy(b[2], "string3");
    strcpy(b[3], "string4");
    strcpy(b[4], "string5");

    /*

       b[]              0   1   2   3   4   5   6    7     8   9
    +--------+        +---+---+---+---+---+---+---+------+---+---+
    |      --|------->| s | t | r | i | n | g | 1 | '\0' |   |   |
    +--------+        +---+---+---+---+---+---+---+------+---+---+
    |      --|------->| s | t | r | i | n | g | 2 | '\0' |   |   |
    +--------+        +---+---+---+---+---+---+---+------+---+---+
    |      --|------->| s | t | r | i | n | g | 3 | '\0' |   |   |
    +--------+        +---+---+---+---+---+---+---+------+---+---+
    |      --|------->| s | t | r | i | n | g | 4 | '\0' |   |   |
    +--------+        +---+---+---+---+---+---+---+------+---+---+
    |      --|------->| s | t | r | i | n | g | 5 | '\0' |   |   |
    +--------+        +---+---+---+---+---+---+---+------+---+---+

    */

    char (*ptr1)[COL] = a;
    printf("Contents of first array \n");
    for (i=0; i<ROW; i++)
        printf("%s \n", *ptr1++);


    char **ptr2 = b;
    printf("Contents of second array \n");
    for (i=0; i<ROW; i++)
        printf("%s \n", ptr2[i]);

    /* b should be free'd */
    for (i=0 ; i<5; i++)
        free(b[i]);

    return 0;
}

Here

char *names [] = {"hello", "Jordan"};

names is array of char pointers i.e it can holds pointers i.e names each elements itself is one char array. But here

char names [] = {"hello", "Jordan"};

names is just a char array i.e it can hold only single char array like "hello" not multiple.

In second case like

int main(void) {
        char names[] = {"hello", "Jordan"};
        return 0;
}

when you compile(Suggest you to compile with -Wall -pedantic -Wstrict-prototypes -Werror flags), compiler clearly says

error: excess elements in char array initializer

which means you can't have more than one char array in this case. Correct one is

char names[] = {'h','e','l','l','o','\0'}; /* here names is array of characters */

Edit :- Also there is more possibility if syntax of names looks like below

char names[] = { "hello" "Jordan" }; /* its a valid one */

then here both hello and Jordan gets joined & it becomes single char array helloJordan.

char names[] = { "helloJordan" }; 

from cdecl:

declare foo as array of pointer to array 2 of pointer to char

char *(*foo[])[2];

So, foo[0] is a pointer to array 2 of char *

That is the array, but for your use, you want:

declare foo as pointer to array 2 of pointer to char;

char *(*foo)[2];

Now you can do:

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

int main() {
    char *(*foo)[2];

    printf("How many people?\n");
    int n; scanf("%d", &n);

    foo = malloc(sizeof *foo * n);

    for (int i = 0; i < n; i++) {
        char bufFirstName[1024];
        char bufLastName[1024];

        printf("Please insert the #%d first and last name:\n", i+1);

        scanf("%s %s", bufFirstName, bufLastName); 

        char *firstName = malloc(strlen(bufFirstName) + 1);
        char *lastName = malloc(strlen(bufLastName) + 1);

        strcpy(firstName, bufFirstName);
        strcpy(lastName, bufLastName);

        foo[i][0] = firstName;
        foo[i][1] = lastName;
    }

    for (int i = 0; i < n; i++) {
        printf("Name: %s LastName: %s\n", foo[i][0], foo[i][1]);
    }

    return 0;
}

Compile with -std=c99

Note that using scanf, strcpy, strlen like that is unsafe because there can be a buffer overflow.

Also, remember to free your malloc's!

Make sure you reserve room for the characters of the strings:

char p[5][128];

and also make sure you limit the length when reading, so scanf() doesn't write outside the buffer:

if(scanf("%127s", p[i]) == 1)
{
  p[i][127] = '\0'; /* Make sure it's terminated. */
}