Step 1: Set up the two checks in a different order.
Check time behaviour first: replace $t$ by $t-t_0$ in the input and see if the output simply slides by $t_0$. Check nonlinearity second: look for any squaring, absolute value, or other operation that breaks proportionality.
Step 2: Look at (M) $y=x^2(t)$.
Delaying the input to $x(t-t_0)$ gives output $x^2(t-t_0)$, which is just the old output shifted by $t_0$, so this one keeps its shape in time: time-invariant. But squaring mixes different inputs together, since scaling the input by $2$ scales the output by $4$, not $2$, so it is non-linear. That is (IV).
Step 3: Look at (N) $y=x(-t)$.
Scaling the input by constants carries straight through the flip, so this rule is linear. For timing, feed in $x(t-t_0)$: the output becomes $x(-t-t_0)$, while a simple delay of the untouched output would give $x(-t+t_0)$. These do not match for $t_0 \ne 0$, so the system's behaviour depends on when the signal arrives: time-varying. That is (III).
Step 4: Look at (O) $y=|x(3-t)|$.
The reflection $3-t$ makes the timing behave like case N above, so this is time-varying. The absolute value on top of that breaks the additive property, since opposite-signed inputs can cancel before the bars but not after, so it is also non-linear. That is (II).
Step 5: Look at (P) $y=x(t-3)$.
This is a plain fixed delay. Scaling and adding inputs passes straight through a delay, so it is linear, and delaying the input further just adds to the existing $3$-second delay, so it is time-invariant. That is (I).
Step 6: Put the table together.
M pairs with IV, N pairs with III, O pairs with II, and P pairs with I. Checking the answer choices, this is option (A).
\[ \boxed{\text{M-IV, N-III, O-II, P-I}} \]
