P(heads) = P
P(tails) = 1-P
Coin is tossed even no of times means 2,4,6,8 .......
For any throw first head should appear after that tails on rest of the throws in individual trial.
P(E) = P(H)P(T) + P(H)P(T)P(T)P(T) + P(H)P(T)P(T)P(T)P(T)P(T)......
P(E)=$P(1-P) + P(1-P)^{3} + P(1-P)^{5} + ........ ( infinite\ GP)$
P(E) = $\frac{P(1-P)}{1-(1-P)^{2}} = \frac{P(1-P)}{(1-1+P)(1+1-P)} = \frac{1-P}{2-P}$
Hence option b) is correct