Let x denote the test statistic for a hypothesis test and $x_0$ its observed value. Then the P-value of the hypothesis test equals

Part (a): $P\left(x\ge x_0\right)$for a right-tailed test

Part (b): $P\left(x\le x_0\right)$for a left-tailed test

Part (c): $2.\min \left\{P\left(x\le x_0\right),P\left(x\ge x_0\right)\right\}$for a two-tailed test

where the probabilities are computed under the assumption.

Consider the given question,

$P\left(x\ge x_0\right)$

Consider a one-mean z-test.

The test statistic for a one-mean z-test using null hypothesis$H_0:\mu ={\mu }_0$is given below,

$z=\frac{\overline{x}-{\mu }_0}{{\displaystyle \raisebox{1ex}{$a$}\!\left/ \!\raisebox{-1ex}{$\sqrt{n}$}\right.}}$

In case the null hypothesis is true then the test statistic will have the standard normal distribution, which is $z~N\left(0,1\right)$.

Let $z_0$ be the observed value of the test statistic z.

In case of the right tailed test, $P-value=P\left(z\ge z_0\right)$, where $z~\hspace{0.17em}N\left(0,1\right)$.

So clearly, the expression is equivalent to the given expression as$$\begin{gathered} x\equiv z, \\ {x}_0\equiv z_0 \end{gathered}$$.

Consider a one-mean z-test.

The test statistic for a one-mean z-test using null hypothesis$H_0:\mu ={\mu }_0$is given below,

$z=\frac{\overline{x}-{\mu }_0}{{\displaystyle \raisebox{1ex}{$a$}\!\left/ \!\raisebox{-1ex}{$\sqrt{n}$}\right.}}$

Let $z_0$ be the observed value of the test statistic z.

In case of the left tailed test, $P-value=P\left(z\le z_0\right)$, where $z~\hspace{0.17em}N\left(0,1\right)$.

So clearly, the expression is equivalent to the given expression as$$\begin{gathered} x\equiv z, \\ {x}_0\equiv z_0 \end{gathered}$$.

Consider a one-mean z-test.

The test statistic for a one-mean z-test using null hypothesis$H_0:\mu ={\mu }_0$is given below,

$z=\frac{\overline{x}-{\mu }_0}{{\displaystyle \raisebox{1ex}{$a$}\!\left/ \!\raisebox{-1ex}{$\sqrt{n}$}\right.}}$

Let $z_0$ be the observed value of the test statistic z.

In case of the two tailed test,

$$\begin{gathered} P-value=2\times min\left\{P\left(z\le z_0\right),P\left(z\ge z_0\right)\right\}......\left(i\right) \\ =\left\{\begin{array}{ll}2\times P\left(z\le z_0\right),& if{z}_0<0\\ or& \\ 2\times P\left(z\ge z_0\right),& if{z}_0>0\end{array}\right.......\left(ii\right) \\ =\left\{\begin{array}{ll}2\times P\left(z\le \left|z_0\right|\right),& if{z}_0<0\\ or& \\ 2\times P\left(z\ge \left|z_0\right|\right),& if{z}_0>0\end{array}\right.\left[\because if{z}_0<0,z_0=-\left|z_0\right|, \\ if{z}_0>0,z_0=\left|z_0\right|\right] \\ =\left\{\begin{array}{ll}2\times P\left(z\le \left|z_0\right|\right),& \\ or& foranyvalueof{z}_0\\ 2\times P\left(z\ge \left|z_0\right|\right)& \end{array}\right. \end{gathered}$$

We can see that the expression of probability given in part (c) is equivalent to the expression of P-value that is obtained.

The justification of step (i) and (ii) is given below,