If $$P = \begin{bmatrix} 1 & 0 \\ \frac{1}{2} & 1 \end{bmatrix}$$, then $$P^{50}$$ is:

We start with the given matrix

$$P=\begin{bmatrix}1&0\\\dfrac12&1\end{bmatrix}.$$

First we separate it into the sum of the identity matrix and another matrix that has zeros on its diagonal. The $$2\times2$$ identity matrix is

$$I=\begin{bmatrix}1&0\\0&1\end{bmatrix}.$$

Comparing $$P$$ with $$I$$, we notice that the only entry that differs is the left-bottom entry $$\dfrac12$$. So we write

$$P=I+N,$$

where

$$N=\begin{bmatrix}0&0\\\dfrac12&0\end{bmatrix}.$$

Now we examine the powers of $$N$$. We multiply $$N$$ by itself:

$$N^2 =\begin{bmatrix}0&0\\\dfrac12&0\end{bmatrix} \begin{bmatrix}0&0\\\dfrac12&0\end{bmatrix} =\begin{bmatrix} 0\cdot0+0\cdot\dfrac12 & 0\cdot0+0\cdot0\\ \dfrac12\cdot0+0\cdot\dfrac12 & \dfrac12\cdot0+0\cdot0 \end{bmatrix} =\begin{bmatrix}0&0\\0&0\end{bmatrix}=0.$$

So $$N$$ is nilpotent of index 2; that is, $$N^2=0$$ and higher powers are also zero:

$$N^3=N^2\cdot N=0,\qquad N^4=0,\quad\text{and so on.}$$

Because $$N^2=0$$, the binomial expansion of $$(I+N)^m$$ truncates after the second term. The general binomial theorem for matrices tells us

$$(I+N)^m=\displaystyle\sum_{k=0}^{m}\binom{m}{k}I^{\,m-k}N^{k}.$$

But since $$N^{2}=0$$, every term with $$k\ge2$$ is zero. So we keep only the $$k=0$$ and $$k=1$$ terms:

$$(I+N)^m=I+\binom{m}{1}N =I+mN.$$

We need the specific case $$m=50$$, therefore

$$P^{50}=(I+N)^{50}=I+50N.$$

Now we substitute $$N=\begin{bmatrix}0&0\\\dfrac12&0\end{bmatrix}$$:

$$50N =50\begin{bmatrix}0&0\\\dfrac12&0\end{bmatrix} =\begin{bmatrix}50\cdot0 & 50\cdot0\\ 50\cdot\dfrac12 & 50\cdot0\end{bmatrix} =\begin{bmatrix}0&0\\25&0\end{bmatrix}.$$

Adding this to the identity matrix gives

$$P^{50}=I+50N =\begin{bmatrix}1&0\\0&1\end{bmatrix} +\begin{bmatrix}0&0\\25&0\end{bmatrix} =\begin{bmatrix}1&0\\25&1\end{bmatrix}.$$

We compare this with the options. It coincides exactly with

$$\begin{bmatrix}1&0\\25&1\end{bmatrix},$$

which is Option A.

Hence, the correct answer is Option A.