Explanation

(a) There are four beats between P and 𝒬, therefore the possible frequencies of P or 254 (that is $250 \pm 4$)Hz. When the prong of P is field, its frequency become greater than the original frequency.
If we assume that the original frequency of P is 254, then on filing its frequency will be greater than 254. The beats between P and 𝒬 will be more than 4. But it is given that the beats are reduced to 2, therefore, 254 is not possible. Therefore, the required frequency must be 246 Hz.