Let the $4$ tosses be named $P,Q,R$ and $S$
To have $3$ consecutive heads:
- $Q$, $R$ must be both heads.
- At least one of $\left \{ P, S \right \}$ must be a head.
Thus, the probability of getting 3 consecutive heads is given by:
$P = P_q \times P_r \times \underbrace{\left ( P_p + P_s - P_p P_s \right )}_{\text{atleast one}}$
$\quad= \dfrac 12 \times \dfrac 12 \times \left ( \dfrac 12 + \dfrac 12 - \dfrac 12 \cdot \dfrac 12 \right )$
$ P = \frac{3}{16}.$
Hence, option D is the correct answer.
Another way of looking at it is:
$P = P_{HHHT} + P_{THHH} + P_{HHHH} = \dfrac{1}{16} + \dfrac{1}{16} + \dfrac{1}{16} = \dfrac{3}{16}$