A PAM signal is transmitted with bit 1 represented by $\pm A$ and bit 0 by 0 for a duration of T. They asked me to get the probability of error of the signaling, I found that $P_{e}=\frac 32*Q(\frac{A}{\sigma})$. But the result given is: $P_{e}=Q(\frac{3A}{2\sigma})+2Q(\frac{A}{2\sigma})$
Did I make a mistake when calculating the probability of error for bit 1?
I took the Threshold of decision is $\pm \frac A2$ then
$P_{e}(bit 1)=\frac 14 *(Q(\frac{-A/2+A}{\sigma})+1-Q(\frac{A/2-A}{\sigma}))$
Both your calculation and your "book"'s answer are incorrect.
When a $1$ is transmitted (whether as $+A$ or $-A$, it doesn't matter), the receiver decision is incorrect if and only if the demodulator output is in the interval $\left[-\frac A2, +\frac A2\right]$. Hence, $$P_{e,1} = Q\left(\frac{A}{2\sigma}\right) - Q\left(\frac{3A}{2\sigma}\right).$$ When a $0$ is transmitted, the receiver decision is incorrect if and only if the demodulator output is not in the interval $\left[-\frac A2, +\frac A2\right]$. Hence, $$P_{e,0} = 2Q\left(\frac{A}{2\sigma}\right).$$ Assuming $0$ and $1$ are equally likely to be transmitted, the average error probability is $$P_e = \frac 32 Q\left(\frac{A}{2\sigma}\right) - \frac 12 Q\left(\frac{3A}{2\sigma}\right).$$ If the transmitter were not choosing randomly between $\pm A$ when transmitting a $1$, the threshold would be $\frac A2$ ($-\frac A2$ for negative thinkers) and a smaller error probability $Q\left(\frac{A}{2\sigma}\right)$ would be achieved.
