In the task we get a lengthly description. Short version is:

1. We have ECC encryption.
2. The curve is y^2 = x^3 + 658974*x + B with most significat decimal digit of B being 6
3. Prime field is based on p = 962280654317
4. We get generator point P = (518459267012, 339109212996)
5. We get a set of points Qi where Qi = ki * P
6. The final secret value is a combination of ki values (this was very unclear, but in the end it was just concatenation of ki values treated as decimal strings and this converted to bytes).

We start-off by recovering the missing B parameter of the curve. This is trivial, since we've got points on the curve and we can just solve the equation y^2 = x^3 + 658974*x + B. We transform this to:

B = (y^2 - x^3 - 658974*x) mod p

And apply this to P = (518459267012, 339109212996) which gives us 618, and it fits the hint from the description that 6 is the most significant digit of B.

prime = 962280654317
x = 518459267012
y = 339109212996
A = 658974
B = (y^2 - x^3 - A*x) %prime

Now we need to recover the ki values. The prime field is very small, so we can efficiently calculate so-called discrete logarithm over the elliptic curve. Fortunately sage has all of this already implemented so we just need to do:

K.<z> = GF(prime)
E = EllipticCurve(K,[A,B])
P = E([x,y]) # generator point

solutions = []
for px,py in data:
    Q = E([px,py])
    solution = P.discrete_log(Q)
    print(px,py, solution)
    solutions.append(solution)
print(solutions)

And from this we recover the ki values. We combine them to get the secret K:

large_string = "".join([str(ki) for ki in solutions])
print("".join(map(lambda x: chr(int(x)), chunk(large_string, 2))))

This gives:

CONVERT THIS TO LOWER CASE FIRST :
THIS IMAGE CONTAINS THE FLAG, TRY TO GET IT
THE SUBMITTED FLAG MUST BE IN THIS FORMAT: 
FLAG-EC[WHAT YOU'LL FIND IN THE IMAGE]
IMAGE URL:
HTTP://CRYPTO.CTFSECURINETS.COM/1/STEG-PART.PNG

Fortunately there is not much of a stegano there really. The picture is:

And can be easily read to get the final flag: flag-ec[EC_St!e-g1(a)no]