Consider a population with the pdf $N(\theta,1)$ where $\theta$ is unknown and hypothesis $H_0:\theta=5.5$ and $H_1:\theta=8$.Suppose that $\bar{X}=\frac{1}{9}\sum_{i=1}^9 X_i.$Reject $H_0$ iff $\bar{X} >7.5$.Power function is given by $Q(\theta)=P_{\theta }(\cal R).$ $\beta=P{\text{(type-II error)}}=P({\text{Accepting } H_0}$ when $H_1 \text{is true}).$
Why then $Q(\theta_1)=1-\beta$ and $Q(\theta)=1-\Phi(22.5-3\theta)?$
In particular how arises $3$ in front of the $\theta$?
You have some undefined non-standard notation in your question. I gather that the power is $P(\bar X \ge 7.5 | \theta = 8) \approx 0.9332.$
If so, in order to evaluate the probability you can use: $$P(\bar X > 7.5) = P\left(\frac {\bar X - \theta}{\sigma/\sqrt{n}} > \frac{7.5 - 8}{1/3} \right) = P(Z > -1.5) = 0.9332,$$ where the result is obtained from printed tables of the standard normal CDF.
Alternatively, you could use software or a statistical calculator. From R statistical software we get:
Below are graphs of the densities of $\bar X$ (based on $n = 9$ observations) for $H_0: \theta = 5.5$ (blue) and $H_a: \theta = 8$ (brown). The power is the probability under the brown curve to the right on the vertical line at $7.5.$ A better choice for the critical value would have been something near $6.75$: then probabilities of both Type I and Type II error would be essentially $0.$
Note: You asked about the
3: I believe it is the square root of the sample size.