The key point is that Slutsky's theorem delivers convergence in distribution, not in probability, whenever the limit involves the weakly-convergent sequence $Y_n$ .

(B) is true. Simplify first:

$\frac{n(X_n-4)+Y_n}{n}=(X_n-4)+\frac{Y_n}{n}.$

Now $X_n\overset{P}{\longrightarrow}3$ gives $X_n-4\overset{P}{\longrightarrow}-1$ . And $Y_n\overset{d}{\longrightarrow} Y$ makes $\{Y_n\}$ bounded in probability ( $O_P(1)$ ), so $Y_n/n\overset{P}{\longrightarrow}0$ . The sum of a sequence converging in probability to $-1$ and one converging in probability to $0$ converges in probability to $-1$ . Note that here the limit is the constant $-1$ , which is why convergence in probability is legitimate.

(A) is false. Slutsky gives $X_n+Y_n\overset{d}{\longrightarrow}3+Y$ , convergence in distribution only. Convergence in probability to the random variable $3+Y$ would require $Y_n$ itself to converge in probability, which is not assumed.

(C) is false for the same reason: $X_nY_n+Y_n=(X_n+1)Y_n\overset{d}{\longrightarrow}4Y$ in distribution, not in probability.

(D) is false. Here

$\frac{n(Y_n-4)+X_n}{n}=(Y_n-4)+\frac{X_n}{n} \overset{d}{\longrightarrow} Y-4 ,$

a non-degenerate random limit — not the constant $-4$ , and not in probability.