Note what is being asked — the incorrect statements. Two of the four are deliberate misstatements of standard results.

Statement (a) — INCORRECT. In frequency modulation the instantaneous carrier frequency follows the amplitude of the modulating signal, not its frequency:

\(f_i(t)=f_c+k_f\,m(t)\)

The modulating frequency controls how rapidly the carrier is swung, but the size of the swing — the deviation ∆f — is set purely by the message amplitude. Swapping "amplitude" for "frequency" here is the classic FM trap.

Statement (c) — INCORRECT. PCM does not encode the sampled amplitude directly. The chain is sample → quantise → encode. The quantiser rounds each sample to one of a finite set of levels, and only that rounded value is turned into a binary code word. Skipping quantisation is precisely what distinguishes PCM from PAM, so this statement actually describes PAM.

Statement (b) — correct, so not an answer. PAM transmits information in short, narrow pulses; to deliver the same average power in a small duty cycle the pulse peaks must be high, so the peak power does exceed that of conventional AM. This is a well-known drawback of pulse-amplitude modulation.

Statement (d) — correct, so not an answer. With L = 2ⁿ levels over a range 2V, the step size is \(\Delta=2V/L\) and the quantisation noise power is

\(N_q=\dfrac{\Delta^{2}}{12}\)

Doubling the number of levels halves ∆ and cuts the noise power by four, giving the familiar rule that each extra bit buys about 6 dB of signal-to-noise ratio:

\(\left(\dfrac{S}{N}\right)_{dB}\approx 1.76+6.02n\)

Step — assemble. The incorrect statements are (a) and (c), which is option 4.

The price of extra levels. More quantising levels mean more bits per sample, hence a higher bit rate \(R=nf_s\) and a wider channel bandwidth — the standard bandwidth-versus-noise trade-off that PCM makes explicit.

Hence, the incorrect statements are (a) and (c).