Let $$p(x)$$ be a quadratic polynomial such that $$p(0) = 1$$. If $$p(x)$$ leaves remainder 4 when divided by $$x - 1$$ and it leaves remainder 6 when divided by $$x + 1$$ then:

We are told that $$p(x)$$ is a quadratic polynomial, so we may write it in its general form as

$$p(x)=ax^{2}+bx+c.$$

First, we use the given value $$p(0)=1.$$ Substituting $$x=0$$ into the general form gives

$$p(0)=a(0)^{2}+b(0)+c=c.$$

But $$p(0)=1$$, therefore

$$c=1.$$

Next, we employ the Remainder Theorem, which states that when a polynomial $$f(x)$$ is divided by $$x-k$$, the remainder is simply $$f(k).$$ We have two such pieces of information.

When $$p(x)$$ is divided by $$x-1$$ the remainder is $$4$$, so

$$p(1)=4.$$

Substituting $$x=1$$ in $$p(x)=ax^{2}+bx+1$$ (remember $$c=1$$) we obtain

$$a(1)^{2}+b(1)+1=4,$$

which simplifies to

$$a+b+1=4.$$

Subtracting $$1$$ from both sides gives

$$a+b=3.\qquad(1)$$

Similarly, when $$p(x)$$ is divided by $$x+1$$ the remainder is $$6$$, so

$$p(-1)=6.$$

Substituting $$x=-1$$ into $$p(x)=ax^{2}+bx+1$$ yields

$$a(-1)^{2}+b(-1)+1=6,$$

that is

$$a-b+1=6.$$

Subtracting $$1$$ from both sides we get

$$a-b=5.\qquad(2)$$

We now have the pair of linear equations

$$a+b=3\quad\text{and}\quad a-b=5.$$

Adding equations (1) and (2) eliminates $$b$$:

$$\bigl(a+b\bigr)+\bigl(a-b\bigr)=3+5\;\Longrightarrow\;2a=8,$$

so

$$a=4.$$

Substituting $$a=4$$ back into equation (1) gives

$$4+b=3\;\Longrightarrow\;b=3-4=-1.$$

Thus the complete quadratic polynomial is

$$p(x)=4x^{2}-x+1.$$

We now evaluate this polynomial at the two points mentioned in the options.

First, for $$x=2$$:

$$p(2)=4(2)^{2}-2+1=4\cdot4-2+1=16-2+1=15.$$

Second, for $$x=-2$$:

$$p(-2)=4(-2)^{2}-(-2)+1=4\cdot4+2+1=16+2+1=19.$$

So we have

$$p(2)=15\quad\text{and}\quad p(-2)=19.$$

Looking at the four options, only the statement $$p(-2)=19$$ is correct.

Hence, the correct answer is Option A.