Doubly Linked List in C, showing head pointer values instead of inserted values

struct Node* NewNode = (struct Node*)(malloc(sizeof(struct Node*)));

First step is to remove some excess parentheses!

struct Node* NewNode = (struct Node*)malloc( sizeof(struct Node*));

Less code makes it easier to spot what may not be correct... So let's fix that.

struct Node* NewNode = (struct Node*)malloc(sizeof(struct Node));

You probably shouldn't be "casting" the return from malloc( ) unless you're using a very old compiler.

struct Node* NewNode = malloc(sizeof(struct Node));

struct Node is repeated, and may create a hard-to-find bug if the code is altered carelessly. How much space do we want? Enough to point at with our pointer!

struct Node* NewNode = malloc( sizeof *NewNode );

Easier to read?

And, wars have been fought over this, but this example shows the benefit of shifting the '*' in the declaration of a pointer.

struct Node *NewNode = malloc( sizeof *NewNode );

NewNode is first-and-foremost a pointer. It just happens to be used to point an instance of a "struct Node".

When the struct is declared, one can save even more repetition by using a typedef that need not be combined with the declaration

struct foo {
    int bar0;
    int bar1;
};
typedef struct foo foo_t; // a conventional naming style
/* or  combined...*/
typedef struct foo {
    int bar0;
    int bar1;
} foo_t;

Had a typedef been used, the following would be even more terse

Node_t *NewNode = malloc( sizeof *NewNode );

As mentioned in the comments, you should allocate memory as Node not Node*.

struct Node* NewNode = (struct Node*) malloc(sizeof(struct Node));

It allocates enough memory for a Node structure and returns a pointer to that allocated memory. You can take a look at this reference.

you have only one small problem which is this line

struct Node* NewNode = (struct Node*)(malloc(sizeof(struct Node*)));

it's supposed to be like this struct Node* NewNode = (struct Node*)(malloc(sizeof(struct Node)));

as sizeof(struct Node*) will return the size of a pointer and the size of any type of pointer in c is mostly 4 bytes , but the size of Node is actually 12 bytes not 4 bytes as

int data; -> 4 bytes

struct Node* next; -> 4bytes

struct Node* prev; -> 4bytes

,so the total is 12 bytes , actually if you write : struct Node* NewNode = (struct Node*)(malloc(12)); will actually work, that's what made the error. if the total size of elements of struct called Node was 4 bytes like it only has one member whose size is 4 bytes then the statement sizeof(struct Node) will have same effect as sizeof(struct Node*) which is a correct answer.

and here is the full code edited with image of the output :

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

struct Node {
    int data;
    struct Node* next;
    struct Node* prev;
};

struct Node* head;

void InsertatBegin(int a) {
    struct Node* NewNode = (struct Node*)(malloc(sizeof(struct Node)));
    NewNode->data = a;
    NewNode->next = NULL;
    NewNode->prev = NULL;
    if (head == NULL) {
        head = NewNode;
        return;
    }
    NewNode->next = head;
    head->prev = NewNode;
    head = NewNode;

}

void traverse() {
    struct Node* temp = head;
    while (temp != NULL) {
        if (temp->next == NULL) {
            printf(" %d->NULL", temp->data);
        }
        else {
            printf(" %d->", temp->data);
        }
        temp = temp->next; // Traversing the List till end
    }
    printf("\n");
}

int main() {
    head = NULL;
    InsertatBegin(5);
    InsertatBegin(6);
    InsertatBegin(7);
    InsertatBegin(8);
    InsertatBegin(9);
    traverse();

}