The given problem involves four voice signals, each limited to 4 kHz, and they are sampled at the Nyquist rate. These signals are then converted into binary PCM (Pulse Code Modulation) signals using 256 quantization levels. We need to determine the bit transmission rate for the time-division multiplexed signal.

To solve this problem, let's break it down into steps:

1. Nyquist Rate:
The Nyquist rate states that the sampling frequency should be at least twice the highest frequency component in the signal to avoid aliasing. In this case, each voice signal is limited to 4 kHz. Therefore, the Nyquist rate for each voice signal is given by:

Nyquist rate = 2 × 4 kHz = 8 kHz

2. Number of Quantization Levels:
The PCM signal is converted using 256 quantization levels. This means that each voice signal is divided into 256 discrete levels to represent the analog signal digitally.

3. Bit Transmission Rate for a Single Voice Signal:
To calculate the bit transmission rate for a single voice signal, we need to consider the Nyquist rate and the number of quantization levels.

Bit transmission rate = Nyquist rate × Number of quantization levels

For each voice signal:
Bit transmission rate = 8 kHz × 256 = 2.048 Mbps

4. Time-Division Multiplexing:
In time-division multiplexing, multiple signals are combined into a single signal by allocating specific time slots to each signal. In this case, we have four voice signals being multiplexed.

Since the bit transmission rate for each voice signal is 2.048 Mbps, the total bit transmission rate for the time-division multiplexed signal is given by:

Total bit transmission rate = Bit transmission rate per voice signal × Number of voice signals

Total bit transmission rate = 2.048 Mbps × 4 = 8.192 Mbps

5. Conversion to kbps:
Finally, we convert the total bit transmission rate to kbps by dividing it by 1000.

Total bit transmission rate in kbps = 8.192 Mbps / 1000 = 8.192 kbps

Therefore, the correct answer is option 'C' - 256 kbps.