solution of  the equation $a^2+pb^2-2c^2-2kcd+(p+k^2)d^2=0$ i am wondering if there is a complete solution for the equation $a^2+pb^2-2c^2-2kcd+(p+k^2)d^2=0$ in which $a,b,c,d,k$ are integer(not all zero) and $p$ is odd prime. 

I'm not getting much 2-adic information for this one, but it should be easy enough to check all solutions mod 8 and mod 16 and see what happens.
To restrict anything, one property requires $p \equiv \pm 3 \pmod 8$ and the other requires   $p \equiv  3 \pmod 4.$ Put them together, when $$p \equiv  3 \pmod 8  $$  and
$$  p | k, $$
then all four of your letters $$ a,b,c,d = 0.$$
The proof uses two flavors of anisotropy for binaries. The assumption is that at least one of
$ a,b,c,d $ is nonzero and  $\gcd(a,b,c,d) = 1.$
 First we have forced 
$a^2 - 2 c^2 \equiv 0 \pmod p,$ so $a,c \equiv 0 \pmod p$ as $(2 | p) = -1.$ But then
$ p b^2 + p d^2 \equiv 0 \pmod {p^2},$ or
 $  b^2 +  d^2 \equiv 0 \pmod p,$  so $b,d \equiv 0 \pmod p$ as $(-1 | p) = -1.$ So 
 $ p | \gcd(a,b,c,d)$ contrary to assumption.
Otherwise, given a fixed $(p,k)$ once you have a nontrivial solution you get infinitely many
using automorphs of the indefinite part in variables $(c,d).$ That is, there may be many parametrized families of solutions of one type or another. But you can figure some of those out with a computer algebra system more easily than I can by hand.
The next interesting case is when $12 k^2 + 8p$ is a square, which means that the binary form 
$T(c,d)=2c^2+2kcd-(p+k^2)d^2$ factors. So $3 k^2 + 2p$ is a square, which is not possible for even $k,$ so $k$ is odd and $2p \equiv 6 \pmod 8,$ or   $p \equiv -1 \pmod 4.$ Unless $p=3$ we also need 
  $p \equiv -1 \pmod 3,$ or   $p \equiv -1  \equiv 11  \pmod {12}.$
For example, with $p=11, k=1,3 k^2 + 2p = 25, p + k^2 = 12,$ we have
$$ a^2+11b^2-2c^2-2cd+12d^2 = a^2+11b^2-2(c-2d)(c+3d).$$
The value of the factorization is that we can take, for instance, $c = 2 d + 1, c + 3 d = 5 d + 1,$ and
$$ a^2+11b^2-2(5d+1) = 0.$$ Now $a^2 + 11 b^2$ is not even unless it is also 
divisible by $4.$ We also need $ a^2 \equiv b^2 \equiv 1 \pmod 5.$ Put them together, we have a parametrized solution of sorts, with $$  a \equiv 1,4 \pmod 5, \; \;   b \equiv 1,4 \pmod 5, \; \;  a \equiv  b \pmod 2$$ take $c = 2 d + 1$ and
$$ d = \frac{ a^2+11b^2-2}{10}.$$  
