I've painted my most interest part of this post, with red color. Please scroll down to see it.
$$P_{\text{B}}(x):=\binom{n}{x}p^x q^{n-x}\tag{1}$$
$$ \begin{cases} p+q=1\\ x\in\left\{0,1,2,~. . . ~,n\right\}\\ 1\ll n,x\in\mathbb{N} \end{cases}\tag{2} $$
$$g(x):=\ln\left(P_{\text{B}}(x)\right)\tag{3}$$
$$=\ln\left(\binom{n}{x}p^xq^{n-x}\right)\tag{4}$$
$$=\ln\left(\frac{n!}{\left(n-x\right)!x!}p^x\left(1-p\right)^{n-x}\right)\tag{5}$$
$$=\ln\left(\frac{n!}{\left(n-x\right)!x!}\right)+\ln\left(p^{x}\left(1-p\right)^{n-x}\right)\tag{6}$$
$$=\ln\left(n!\right)-\ln\left(\left(n-x\right)!x!\right)+\ln\left(p^{x}\right)+\ln\left(\left(1-p\right)^{n-x}\right)\tag{7}$$
$$=\ln\left(n!\right)-\left(\ln\left(\left(n-x\right)!\right)+\ln\left(x!\right)\right)+x\ln\left(p\right)+\left(n-x\right)\ln\left(1-p\right)\tag{8}$$
$$=\ln\left(n!\right)-\ln\left(\left(n-x\right)!\right)-\ln\left(x!\right)+\ln\left(p\right)x+\ln\left(1-p\right)\left(n-x\right)\tag{9}$$
$$\frac{\mathrm{d}}{\mathrm{dx}}g(x)\tag{10}$$
$$=\frac{\mathrm{d}}{\mathrm{dx}}\left(\ln\left(n!\right)-\ln\left(\left(n-x\right)!\right)-\ln\left(x!\right)+\ln\left(p\right)x+\ln\left(1-p\right)\left(n-x\right)\right)\tag{11}$$
$$=-\frac{\mathrm{d}}{\mathrm{dx}}\ln\left(\left(n-x\right)!\right)-\frac{\mathrm{d}}{\mathrm{dx}}\ln\left(x!\right)+\ln\left(p\right)\frac{\mathrm{d}}{\mathrm{dx}}x+\ln\left(1-p\right)\frac{\mathrm{d}}{\mathrm{dx}}\left(n-x\right)\tag{12}$$
$$=-\frac{\mathrm{d}}{\mathrm{dx}}\ln\left(\left(n-x\right)!\right)-\frac{\mathrm{d}}{\mathrm{dx}}\ln\left(x!\right)+\ln\left(p\right)-\ln\left(1-p\right)\tag{13}$$
$$\frac{\mathrm{d}}{\mathrm{dx}}\ln\left(x!\right)\approx\ln\left(x\right)~~\leftarrow~~\text{this apporximation is to be used}\tag{14}$$
About the first term of eqn13,
$$u:=n-x\tag{15}$$
$$\frac{\mathrm{d}}{\mathrm{dx}}\left(\ln\left(\left(n-x\right)!\right)\right)\tag{16}$$
$$=\frac{\mathrm{d}}{\mathrm{dx}}\ln\left(u!\right)\tag{17}$$
$$=\frac{\mathrm{d}}{\mathrm{du}}\ln\left(u!\right)\frac{\mathrm{du}}{\mathrm{dx}}\tag{18}$$
$$\approx\ln\left(u\right)\frac{\mathrm{d}}{\mathrm{dx}}\left(n-x\right)\tag{19}$$
$$=-\ln\left(n-x\right)\tag{20}$$
$$\therefore~~\frac{\mathrm{d}}{\mathrm{dx}}g(x)\approx\ln\left(n-x\right)-\ln\left(x\right)+\ln\left(p\right)-\ln\left(1-p\right)\tag{21}$$
$$=\ln\left(\frac{n-x}{x}\right)+\ln\left(\frac{p}{1-p}\right)\tag{22}$$
$$=\ln\left(\frac{\left(n-x\right)}{x}\cdot\frac{p}{1-p}\right)\tag{23}$$
$$x~~\text{is now assumed as continuous variable}\tag{24}$$
$$\text{We will find value(s)of}~~x~~\text{such that}~~\frac{\mathrm{d}}{\mathrm{dx}}g(x)=0\tag{25}$$
$$\frac{\mathrm{d}}{\mathrm{dx}}g(x)=\ln\left(\frac{p\left(n-x\right)}{x\left(1-p\right)}\right)\tag{26}$$
$$\therefore~~\frac{p\left(n-x\right)}{x\left(1-p\right)}=1\tag{27}$$
$$p\left(n-x\right)=x\left(1-p\right)\tag{28}$$
$$np-px=x-xp\tag{29}$$
$$\therefore~~x=np\tag{30}$$
$$\mu_{}:=np\tag{31}$$
Currently I can't get the following claim.
$$\color{red}{P_{\text{B}}(x)~~\text{takes a local maximum value where}~x=np~~\text{is held}}\tag{32}$$
I've enumerated my thoughts for it as follows.
$~P_{\text{B}}(x)~$draws a graph which is resemble with following, as number of iterations$~(=n)~$is large.
This means that local extremum value of$~P_{\text{B}}(x)~$can be immediately determined as local maximum value.
Cited it from here
$$g(x)=\ln\left(P_{\text{B}}(x)\right)\tag{33}$$
About above,$~g(x)~$returns a value of natural log function though argument for it is not linear as same as$~x~$for instance. $~P_{\text{B}}(x)~$can vary a slope itself depends on range where$~x~$belongs.
Since natural log function is a monotonic increasing function, the following claims are true.
$$
\begin{cases}
P_{\text{B}}(x)~~\text{increases}~~\Leftrightarrow~~g(x)~~\text{increases}\\
P_{\text{B}}(x)~~\text{decreases}~~\Leftrightarrow~~g(x)~~\text{decreases}\\
\end{cases}\tag{34}
$$
And still I can't get line32.
$$P= \binom{n}{x}\, p^x\,q^{n-x}=\frac{\Gamma (n+1) }{\Gamma (x+1)\, \Gamma (n-x+1)}p^x\, q^{n-x}$$ Using logarithmic differentiation and replacing the digamma function by generalized harmonic numbers $$\frac 1 P \frac{dP}{dx}=H_{n-x}-H_x+\log (p)-\log (q)$$
Using the approximation $$H_p \simeq \gamma +\log(p)+O\left(\frac{1}{p}\right)$$ then $$x=\frac{n p}{p+q}=np$$