Let the smallest value of $$k \in \mathbb{N}$$, for which the coefficient of $$x^3$$ in $$(1+x)^3 + (1+x)^4 + \ldots + (1+x)^{99} + (1+kx)^{100}$$, $$x \neq 0$$, is $$\left(43n + \frac{101}{4}\right)\binom{100}{3}$$ for some $$n \in \mathbb{N}$$, be $$p$$. Then the value of $$p + n$$ is :

Coefficient of $$x^{3}$$ in $$(1+x)^{m}$$ is $$\binom{m}{3}$$.

Hence the coefficient contributed by the terms $$(1+x)^{3}+(1+x)^{4}+\ldots +(1+x)^{99}$$ equals

$$\sum_{m=3}^{99}\binom{m}{3}.$$

Using the identity $$\sum_{r=0}^{N}\binom{r}{k}=\binom{N+1}{k+1},$$ we get

$$\sum_{m=3}^{99}\binom{m}{3}=\binom{100}{4}.$$

The last term $$(1+kx)^{100}$$ contributes $$\binom{100}{3}k^{3}$$ to the coefficient of $$x^{3}$$.

Therefore the required total coefficient is

$$\binom{100}{4}+\binom{100}{3}k^{3}.$$

According to the question, this must equal $$\left(43n+\frac{101}{4}\right)\binom{100}{3}.$$ Dividing by $$\binom{100}{3}$$ gives

$$\frac{\binom{100}{4}}{\binom{100}{3}}+k^{3}=43n+\frac{101}{4}.$$

Simplify $$\dfrac{\binom{100}{4}}{\binom{100}{3}}$$:

$$\frac{\dfrac{100!}{4!\,96!}}{\dfrac{100!}{3!\,97!}}=\frac{3!\,97!}{4!\,96!}=\frac{6}{24}\times97=\frac{97}{4}.$$

Hence

$$\frac{97}{4}+k^{3}=43n+\frac{101}{4}.$$

Multiplying by $$4$$:

$$97+4k^{3}=172n+101\quad\Longrightarrow\quad4k^{3}=172n+4.$$

Dividing by $$4$$ gives the key relation

$$k^{3}-1=43n.$$

This means $$k^{3}\equiv1\pmod{43}$$ and $$n=\dfrac{k^{3}-1}{43}\in\mathbb{N}.$$ We need the smallest natural $$k$$ giving a positive $$n$$.

Test successive values of $$k$$ (mod $$43$$):

$$k=1\;: \;k^{3}=1\;\;(\Rightarrow n=0,\;{\text{not allowed}})$$
$$k=2\;: \;2^{3}=8\neq1$$
$$k=3\;: \;3^{3}=27\neq1$$
$$k=4\;: \;4^{3}=64\equiv21\ (\text{mod }43)$$
$$k=5\;: \;5^{3}=125\equiv39\ (\text{mod }43)$$
$$k=6\;: \;6^{3}=216\equiv1\ (\text{mod }43).$$

Thus the smallest suitable $$k$$ is $$k=6$$. For this value,

$$n=\frac{6^{3}-1}{43}=\frac{215}{43}=5.$$

Hence $$p=k=6$$ and

$$p+n=6+5=11.$$

Option B which is: $$11$$