1. Let us use total probability based on the coin toss.

• Probability of Head, $P(H)=\dfrac12$
• Probability of Tail, $P(T)=\dfrac12$

We need: $P(\text{noted number is }7\text{ or }8)$

2. Case 1: Coin shows Head

Then two fair dice are rolled, and we note their sum.

Total outcomes for two dice: $36$

Favourable outcomes for sum $7$: $(1,6),(2,5),(3,4),(4,3),(5,2),(6,1)$ So, $6\text{ outcomes}$

Favourable outcomes for sum $8$: $(2,6),(3,5),(4,4),(5,3),(6,2)$ So, $5\text{ outcomes}$

Hence, $P(7\text{ or }8\mid H)=\frac{6+5}{36}=\frac{11}{36}$

3. Case 2: Coin shows Tail

Then one card is chosen from cards numbered $1$ to $9$.

Total cards: $9$

Favourable cards: $7,8$ So, $2\text{ favourable cards}$

Hence, $P(7\text{ or }8\mid T)=\frac{2}{9}$

4. Apply the law of total probability:

$P(7\text{ or }8)=P(H)\,P(7\text{ or }8\mid H)+P(T)\,P(7\text{ or }8\mid T)$

Substitute values: $P=\frac12\cdot\frac{11}{36}+\frac12\cdot\frac{2}{9}$

Convert to common denominator: $\frac12\cdot\frac{2}{9}=\frac{1}{9}=\frac{4}{36}$

So, $P=\frac{11}{72}+\frac{16}{72}=\frac{27}{72}$ Wait carefully — let us recompute properly:

$\frac12\cdot\frac{11}{36}=\frac{11}{72}$ $\frac12\cdot\frac{2}{9}=\frac{1}{9}=\frac{8}{72}$

Therefore, $P=\frac{11}{72}+\frac{8}{72}=\frac{19}{72}$

5. Hence the required probability is $\boxed{\frac{19}{72}}$

So the correct option is D.