The given polynomial can be written in summation form as
$$P(x)=\sum_{k=0}^{1000}(k+1)\,x^{k}\,(1+x)^{1000-k}\qquad -(1)$$
To get the coefficient of $$x^{50}$$ we expand every factor $$ (1+x)^{1000-k} $$ and collect all those products whose total power of $$x$$ becomes 50.
For a fixed $$k$$, the exponent contributed by $$x^{k}$$ is $$k$$.
Hence we need the term $$x^{\,50-k}$$ from the expansion of $$(1+x)^{1000-k}$$.
Its coefficient is $$\binom{1000-k}{\,50-k}$$.
Therefore, the required coefficient (call it $$C_{50}$$) equals
$$\displaystyle C_{50}= \sum_{k=0}^{50}(k+1)\binom{1000-k}{\,50-k}\qquad -(2)$$
(The upper limit is 50 because for $$k\gt 50$$ the power $$50-k$$ becomes negative and such terms do not occur.)
Lemma : For integers $$n\ge r\ge 0$$,
$$\displaystyle \sum_{k=0}^{r}(k+1)\binom{n-k}{\,r-k}=\binom{n+2}{\,r}\qquad -(3)$$
Proof of the Lemma (by induction on $$r$$)
Base $$r=0$$ : LHS $$=(0+1)\binom{n}{0}=1=$$ RHS.
Induction step : Assume the result true for some $$r$$; we prove it for $$r+1$$.
Starting from the left hand side, use Pascal’s identity
$$\binom{a}{b}=\binom{a-1}{b}+\binom{a-1}{b-1}$$:
$$\begin{aligned}
\sum_{k=0}^{r+1}(k+1)\binom{n-k}{\,r+1-k}
&=\sum_{k=0}^{r+1}(k+1)\Bigl[\binom{n-k}{\,r-k}+\binom{n-k-1}{\,r-k}\Bigr]\\
&=\underbrace{\sum_{k=0}^{r+1}(k+1)\binom{n-k}{\,r-k}}_{(A)}
+\underbrace{\sum_{k=0}^{r}(k+1)\binom{n-k-1}{\,r-k}}_{(B)}
\end{aligned}$$
(A) contains $$r+2$$ terms, (B) contains $$r+1$$ terms.
Shift the index in (B) once to compare with (A): set $$j=k+1$$ in (B)
$$\begin{aligned}
(B)&=\sum_{j=1}^{r+1}j\binom{n-j}{\,r+1-j}.
\end{aligned}$$
Now combine (A) and (B): every inner term $$j\binom{n-j}{\,r+1-j}$$ of (B) cancels the same part inside (A), leaving only the first term of (A) and the last term of (A). After cancellation one obtains exactly
$$\binom{n+2}{\,r+1}\;,$$
establishing the induction step. Hence (3) is proved for all $$r$$.
Putting $$n=1000,\; r=50$$ in (3):
$$\sum_{k=0}^{50}(k+1)\binom{1000-k}{\,50-k}=\binom{1002}{50}\qquad -(4)$$
Comparing (2) and (4), the coefficient of $$x^{50}$$ in $$P(x)$$ is
$$C_{50}= \binom{1002}{50}.$$
Hence the correct option is:
Option C which is: $$\displaystyle\binom{1002}{50}$$
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