What's the difference between int *p and int p[] as function parameters?

They're identical — a parameter written int p[] decays to int *p, so the function only ever receives a pointer and loses the array's size.

// These function signatures are EXACTLY the same:
void foo(int *arr);
void foo(int arr[]);
// The 10 is ignored!
void foo(int arr[10]);

Inside the function:

void foo(int arr[]) {
    // Returns sizeof(int*), NOT array size!
    sizeof(arr);
    // arr is just a pointer, array size information is lost
}

That's why you need to pass the size separately:

void process(int *arr, size_t len) {
    for (size_t i = 0; i < len; i++) {
        arr[i] *= 2;
    }
}

int main() {
    int data[] = {1, 2, 3, 4, 5};
    process(data, sizeof(data) / sizeof(data[0]));
}

Exception - pointers to arrays preserve size:

// Pointer to array of 10 ints
void foo(int (*arr)[10]) {
    // 40 bytes! Size preserved
    sizeof(*arr);
}