If $p, q$ are polynomials of $\deg \le n$ with coefficients $a_k, b_k$ respectively, then the coefficients of the product $pq$ may be found in terms of those of $p$ and $q$ as follows:
$$\begin{aligned} p(x)q(x) &= \sum_{i=0}^{n} a_i x^i \sum_{j=0}^{n} b_j x^j = \sum_{i=0}^{n} a_i b_j x^{i+j} \\ &= \sum_{k=0}^{2n} \left( \sum_{i+j=k} a_i b_j \right) x^k = \sum_{k=0}^{2n} \left( \sum_{i=0}^{k} a_i b_{k-i} \right) x^k \end{aligned}$$

To obtain expression $B$, apply this formula with
$$p(x) = (1 + x)^n = \sum_{k=0}^{n} \binom{n}{k} x^k \quad \text{and} \quad q(x) = (1 - x)^n = \sum_{k=0}^{n} (-1)^k \binom{n}{k} x^k.$$
Then
$$\sum_{k=0}^{2n} \left( \sum_{i=0}^{k} (-1)^i \binom{n}{i} \binom{n}{k-i} \right) x^k = p(x)q(x) = (1-x^2)^n = \sum_{r=0}^{n} (-1)^r \binom{n}{r} x^{2r},$$
so that, comparing coefficients, we see that, if $0 \le k \le 2n$,
$$\sum_{i=0}^{k} (-1)^i \binom{n}{i} \binom{n}{k-i} = \begin{cases} 0, & \text{if } k \text{ is odd,} \\ (-1)^{\frac{k}{2}} \binom{n}{\frac{k}{2}}, & \text{if } k \text{ is even.} \end{cases}$$
In particular, taking $k = n$,
$$B = \sum_{k=0}^{n} (-1)^k \binom{n}{k}^2 = \begin{cases} 0, & \text{if } n \text{ is odd,} \\ (-1)^{\frac{n}{2}} \binom{n}{\frac{n}{2}}, & \text{if } n \text{ is even.} \end{cases}$$