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$underline{text{p-adic numbers and p-adic power series}}:$

If $ a,X in mathbb{Q}$, then when the binomial expression $f(X)=(1+X)^a in mathbb{Q} $ both in p-adic field $mathbb{Q}_p$ and real field $mathbb{R}$ ?

Answer:

For example let $X=frac{7}{9}$ and $a=frac{1}{2}$, we see $f(frac{7}{9})=(1+frac{7}{9})^{frac{1}{2}}=left(frac{16}{9} right)^{frac{1}{2}}=pm frac{4}{3}$ and fix the prime $p=7$. Thus in $mathbb{Q}_7$ as well as in $mathbb{R}$, the number $frac{16}{9}$ has square $ pm frac{4}{3}$. But in $ mathbb{Q}_7$, the square root $ pm frac{4}{3} equiv 1 (mod 7)$.

Thus in $mathbb{R}$ the value of $f(frac{7}{9})= frac{4}{3} in mathbb{Q} $ and in $ mathbb{Q}_7$ the value of $f(frac{7}{9})=1 (mod 7) in mathbb{Q}$.

Thus in this case $f(X)=(1+X)^a$ gives rational value both in $mathbb{R}$ and $ mathbb{Q}_p$, for prime $p$.

This was particular case.

Can someone help me with the general case so that $ (1+X)^a in mathbb{Q}$ both in $mathbb{R}$ and $ mathbb{Q}_p$?

Not a complete answer, just a collection of hints and remarks.

The series is

$$displaystyle (1+X)^a = sum_{k=0}^infty binom{a}{k}X^k .$$

Real convergence: If $ain Bbb Z_{ge 0}$, the binomial coefficents become eventually $0$ and this is just a finite sum, hence converges for arbitrary $X$. If $anotin Bbb Z_{ge 0}$, one has to use estimates of the usual absolute value of the binomial coefficients. I know next to nothing about this and just googled a bit. According to http://emis.math.tifr.res.in/journals/JIPAM/images/061_06_JIPAM/061_06.pdf, as soon as $a>-1$, certainly $vert Xvert < 1$ is sufficient (and I have the feeling that this bound is reasonable, if not necessary, in general). the series converges for $vert Xvert < 1$ and diverges for $vert Xvert> 1$, as shown with an argument from complex analysis by reuns in a comment.

Real rationality: Answered in Wojowu's comment on How to find all $x in mathbb{Q}$ and $r in mathbb{Q}$ such that $(1+x)^r$ becomes a rational number?.

$p$-adic convergence: Again if $ain Bbb Z_{ge 0}$, the binomial coefficents become eventually $0$ and this is just a finite sum. For general $anotin Bbb Z_{ge 0}$, now one needs estimates for the $p$-adic absolute value of the binomial coefficients. A good thing is that because we are in an ultrametric, we only need to check whether

$$lim_{kto infty}vert binom{a}{k}X^kvert_p = lim_{kto infty}vert binom{a}{k} vert_p cdot vert Xvert_p^k stackrel{?}=0.$$

Again I have to leave this open in general; in analogy to the real case however, one definitely has (cf. http://www.math.uconn.edu/~kconrad/blurbs/gradnumthy/binomialcoeffpadic.pdf, Definition of $p$-adic $(1+x)^alpha$ via binomial series and log/exp): If $vert avert_p le 1$, then all $vert binom{a}{k}vert_p le 1$, meaning that the series certainly converges for those $X$ with $vert Xvert_p < 1$.

$p$-adic rationality: Here is a subtlety. E.g. look at

$$displaystyle (1+(-frac78))^frac13 = sum_{k=0}^infty binom{1/3}{k}(-7/8)^k .$$

In $Bbb R$, the series converges to $1/2$, which is indeed a cube root (more precisely: the unique positive real cube root) of $1/8$. The series also does converge in $Bbb Q_7$, and also to a (!) cube root of $1/8$, but not to $1/2 in Bbb Q$, rather to the unique one which is $equiv 1$ mod $7$, and that is $zeta cdot 1/2 notin Bbb Q$, where $zeta in Bbb Z_7$ is the primitive third root of unity which is $equiv 2$ mod $7$. (Note that $Bbb Z_7$ contains exactly the sixth roots of unity; the primitive sixth ones are $equiv 3$ resp. $equiv 5$ mod $7$, the primitive third ones are $equiv 2$ resp. $equiv 4$ mod $7$, well and there are $pm 1$).

So in general, even in the case $vert avert_p le 1$ and $vert Xvert_p < 1$ and the real rationality criterion is satisfied, it is not necessarily true that the limit of the $p$-adic series is rational. Rather, it is = (the rational we get from the real consideration) times (some root of unity), so that this product is $equiv 1$ mod $p$. Whether that can be fulfilled by the only rational roots of unity, namely $pm 1$, depends on $X, a$ and $p$ again (in your example in the OP, it can, in my above example, it cannot).

To answer a question in the comments: Sure the values of $f(X)$ evaluated in the two different ways can be identical, e.g. in the trivial case that $a in Bbb N$. Or also, I think, $(X,a,p)=(-63/64, 1/4,3)$. What one needs is that the real rationality criterion is satisfied, so that we have a rational real value of $(1+X)^a$, and the numerator of $(1+X)^a -1$ must be divisible by $p$.