Probability mean,variance and standard deviation formula confusion. I have a confusion in the formula attached.
Why and how are the two formulas equivalent ? sigma in the image is the standard deviation of a distribution...

 A: If you have a discrete random variable $X$ that takes on values $x_i$ with probabilities $p(x_i), 1 \leq i \leq n$, and a function $f(x)$, then the expected value of $f(x)$ is given by
$$
E(f(X)) = \sum_{i=1}^n f(x_i) p(x_i)
$$
That is LOTUS.
Perhaps an example will make it clearer.  Suppose that $X$ takes on values $1, 2, 3$ with probabilities $1/2, 1/3, 1/6$, and $f(x) = x^2$.  Then the expected value of $X^2$ is
$$
E(X^2) = \sum_{i=1}^3 x_i^2 p(x_i) = (1)(1/2)+(4)(1/3)+(9)(1/6) = 10/3
$$
In your case, you have $n$ values, not $3$, and the function is $(X-\mu)^2$, not $X^2$, but otherwise, the application is the same.
A: For a discrete random variable $X$ taking on values 
$x_1, x_2, \ldots,x_n$ with probabilities $p_1, p_2, \ldots, p_n$
respectively, the expected value or expectation of $X$, commonly
denoted as $E[X]$, is defined to be
$$E[X] = \sum_{i=1}^n x_i p_i.\tag{1}$$
(Feel free to replace each $p_i$ with $p(x_i)$ if that feels more
comfortable).
Similarly, the expected value of $Y$, a discrete random variable
taking on values $y_1, y_2, \ldots, y_m$ with probabilities
$q_1, q_2, \ldots, q_m$ is 
$$E[Y] = \sum_{j=1}^m y_j q_j.\tag{2}$$
Now consider a function $g(x)$ of a real variable, e.g. $g(x)=ax+b$
or $g(x) = x^2$ and suppose that $Y$ and $X$ are related as
$Y=g(X)$. What is meant by this is that if one knows the
value of $X$ on any trial (e.g. $X$ took on value $x_3$,
then $Y$ must have taken on value $g(x_3)$. The
standard description for this is that $Y$ is a function of $X$.
Let's discuss this relationship between $X$ and $Y$
a little further. Note that by applying $g(\cdot)$
to all the elements of the set $\{x_1, x_2, \ldots, x_n\}$,
we get $n$ numbers $\{g(x_1), g(x_2), \ldots, g(x_n)\}$.
If all these $g(x_i)$.s are distinct numbers, then
it must be that $m = n$ and 
$$\begin{align}
\{g(x_1), g(x_2), \ldots, g(x_n)\} &= \{y_1, y_2, \ldots, y_n\},\\
\{p_1, p_2, \ldots, p_n\} &= \{q_1, q_2, \ldots, q_n\}.
\end{align}$$
More specifically, if $g(x_3) = y_1$, say, then it is also true
that $q_1 = P\{Y= y_1\} = p_3 = P\{X=x_3\}$
Thus, the sum $\sum_{j=1}^n y_j q_j$ in $(2)$ is the same as the
sum $\sum_{i=1}^n g(x_i)p_i$ except that the the terms might
be arranged in different orders in the two sums. In short, 
in this case, we have the result that
$$E[Y] = \sum_{j=1}^n y_jq_j = \sum_{i=1}^n g(x_i)p_i \tag{3}.$$
What if $\{g(x_1), g(x_2), \ldots, g(x_n)\}$ is not a collection
of $n$ distinct numbers but is instead a multiset (in which the
same number can occur more than once). As an illustrative
example, suppose that $g(x_1)=g(x_2)=g(x_3) = y_1$ and all the
other $g(x_i)$ are distinct. Then, we have that 
$$P\{Y=y_1\} = q_1 = P\left\{X \in \{x_1, x_2, x_3\}\right\}
= p_1+p_2+p_3$$
while $q_2, q_3, \ldots, q_{n-2}$ are the same as $p_4, p_5, \ldots,
p_n$ in some order.
Consequently, 
$$\begin{align}
E[Y] &= \sum_{j=1}^{n-2} y_jq_j\\
&= 
y_1q_1 + \sum_{j=2}^{n-2} y_j q_j\\
&= y_1(p_1+p_2+p_3) + \sum_{i=4}^{n} g(x_i)p_i\\
&=  y_1p_1+y_1p_2+y_1p_3 + \sum_{i=4}^{n} g(x_i)p_i\\
&= g(x_1)p_1 + g(x_2)p_2+g(x_3)p_3 +\sum_{i=4}^{n} g(x_i)p_i\\\
&= \sum_{i=1}^n g(x_i)p_i \tag{4}
\end{align}$$
since $y_1 = g(x_1)=g(x_2)=g(x_3)$.  Obviously, the same kind of
substitution can be used if some other $x_i$'s are mapped onto
the same $y_j$ by the function $g(\cdot)$.

Thus, for a discrete random variable $Y$ that
  happens to equal $g(X)$ where $X$ is another discrete random
  variable, $E[Y]$, the expectation of $Y$, which is defined by $(2)$ can also
  be computed via the formula
  $$E[Y] = E[g(X)] = \sum_{i=1}^n g(x_i)p_i. \tag{5}$$

The result $(5)$ is sometimes referred to (by non-statisticians)
as the law of the unconscious statistician (LOTUS for short)
because some statisticans think that $(5)$ is the definition
of the expectation of $g(X)$, seemingly being unconscious that
$g(X)$ is itself a random variable (called $Y$ above) and so its
expectation is given by $(2)$. Thus, there are thus two different
definitions of $E[Y] = E[g(X)]$, and somewhere someone needs to
say (and prove) that the rights sides of $(2)$ and $(5)$ both give 
the same value for the expectation of $Y = g(X)$. LOTUS does just
that. It is a theorem saying that the right side of $(5)$
happens to equal $E[g(X)]$, and not a definition of $E[g(X)]$.

So, how does all this apply to your question? Well, your book defines
the variance $\sigma^2$ to be the sum $\sum_{i=1}^n (x_i-\mu)^2 p_i$.
On the next line, it thinks of LOTUS and says, "Hey, if I define
$g(x)$ to be the function $(x-\mu)^2$, and the random variable
$Y$ as $Y=g(X) = (X-\mu)^2$, then by LOTUS,
$$E[(X-\mu)^2] = \sum_{i=1}^n (x_i-\mu)^2 p_i$$ and the sum on the right
is exactly what I just defined as $\sigma^2$, the variance of $X$." So,

$\displaystyle \sigma^2 = \operatorname{var}(X) = E[(X-\mu)^2].$

