Binomial Expansion of $$(1+x)^n$$

General term in the expansion of $$\left(x^2+\frac1x\right)^{12}$$ is $$T_{r+1} = \binom{12}{r} (x^2)^{12-r} \left(\frac1x\right)^r$$

$$= \binom{12}{r}x^{24-3r}$$

For the coefficient of

$$x^9,$$

$$24-3r=9$$

$$3r=15$$

$$r=5$$

Hence coefficient of $$x^9$$ is $$\binom{12}{5}$$

$$=792$$

Now coefficient of $$x^5$$ in $$(1+x)^n$$ is $$\binom n5$$

Therefore,

$$\binom n5=792$$

Since

$$\binom{12}{5}=792,$$

we get

$$n=12$$

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: $$\displaystyle\binom{1002}{50}$$

Because the denominators are $$r+1$$ relative to the coefficient $$\binom{n}{r}$$, this indicates an integration approach.
$$(1+x)^n = \binom{n}{0} + \binom{n}{1}x + \binom{n}{2}x^2 + \dots + \binom{n}{n}x^n$$
Integrate both sides with respect to $$x$$ from $$0$$ to $$1$$:
$$\int_{0}^{1} (1+x)^n \,dx = \int_{0}^{1} \left( \binom{n}{0} + \binom{n}{1}x + \binom{n}{2}x^2 + \dots + \binom{n}{n}x^n \right) \,dx$$
Compute the definite integral for both sides:
• Left-Hand Side (LHS):
$$\left[ \frac{(1+x)^{n+1}}{n+1} \right]_{0}^{1} = \frac{2^{n+1}}{n+1} - \frac{1^{n+1}}{n+1} = \frac{2^{n+1} - 1}{n+1}$$
• Right-Hand Side (RHS):
$$\left[ \binom{n}{0}x + \binom{n}{1}\frac{x^2}{2} + \binom{n}{2}\frac{x^3}{3} + \dots + \binom{n}{n}\frac{x^{n+1}}{n+1} \right]_{0}^{1}$$
Substituting the limits $$1$$ and $$0$$ yields:
$$\frac{\binom{n}{0}}{1} + \frac{\binom{n}{1}}{2} + \frac{\binom{n}{2}}{3} + \dots + \frac{\binom{n}{n}}{n+1}$$ = $$\frac{2^{n+1} - 1}{n+1}$$

$$=\ \frac{\ 2^8-1}{8}=\ \frac{\ 255}{8}$$

So, $$p+q=255+8=263$$