In function parameter passing, arrays are passed into the function in the form of pointers, and there will be no call by value. In the function parameter, int arr[4] will degenerate into int *, and the 4 will be lost, so a in the isEqual function is actually just the first address of the array a.
If you want to pass in the array pointer and the size of the array at the same time, you need to use the array length as another formal parameter of the function:
For example:

#include "iostream"

using namespace std;

int isEqual(int a[],int length_a ,int b[],int length_b) {
    cout<<length_b<<length_a<<endl;
    if (length_a != length_b) {
        return 200;
    }
    else {
        for (int i = 0; i < length_a; i++) {
               if (a[i] != b[i]) {
                return 200;
            }
        }
        return 30;
    }

}

int main() {
    int arr1[4] = { 1,2,3,5 };
    int arr2[3] = { 1,2,3 };
    int flag = isEqual(arr1,sizeof(arr1)/sizeof(int),arr2,sizeof(arr2)/sizeof(int));
    cout << flag << endl;

    return 0;
}

Because you calculated the length of the two arrays incorrectly

 #include "iostream"

using namespace std;

int isEqual(int a[], int b[], int length_a, int length_b) {
   
    cout << length_a << length_b << endl;
    if (length_a != length_b) {
        return 200;
    }
    else {
        for (int i = 0; i < length_a; i++) {
            cout << a[i] << b[i] << endl;
            if (a[i] != b[i]) {
                return 200;
            }
        }
        return 30;
    }
}

int main() {
    int arr1[5] = { 2,1,2,3,5 };
    int arr2[3] = { 1,2,3 };
    int length_a = sizeof(arr1) / sizeof(arr1[0]);
    int length_b = sizeof(arr2) / sizeof(arr2[0]);
    int flag = isEqual(arr1, arr2, length_a, length_b);
    cout << flag << endl;

    return 0;
}