Factorise the target expression first.

\(a = a\times b + a = a\,(b+1)\)

So any equivalent statement must multiply a by (b + 1).

Statement (a) — correct. A compound assignment in C expands with the whole right-hand side as one operand:

a *= b + 1;  ≡  a = a * (b + 1);

This is the key rule of the question. The right-hand expression is evaluated completely before the multiplication, exactly as if it were parenthesised, so a *= b + 1 is not a = a*b + 1. Expanding, a*(b+1) = a*b + a ✓.

Statement (c) — correct. a = (b + 1) * a; is the same product written the other way round, and multiplication is commutative ✓.

Statement (b) — wrong. != is the inequality operator, not an assignment. The line (a * b) != (b + a); merely computes a truth value, discards it and leaves a completely unchanged ✗.

Statement (d) — wrong. By the same expansion rule, a *= b + a; means a = a * (b + a), which is \(ab+a^{2}\), not \(ab+a\). The two agree only in the special case a = 1 ✗.

Step — assemble. (a) and (c) are equivalent, giving option 4.

Check with numbers. Take a = 3, b = 4. The target is \(3\times4+3=15\). Statement (a): 3 * (4+1) = 15 ✓. Statement (c): (4+1) * 3 = 15 ✓. Statement (d): 3 * (4+3) = 21 ✗. One substitution settles the whole question.

The general rule to carry away. For any compound operator, x op= expr is defined as x = x op (expr) — the implied parentheses around the entire right-hand side are what trip people up with *=, /= and -=. Note also that x op= expr evaluates x only once, which matters when the left side has side effects such as arr[i++] *= 2.

Hence, the equivalent forms are (a) and (c).