With a sample as large as $n = 1000$ and $p$ in the vicinity of 1/2,
a normal approximation should work well.
To test the null hypothesis
$H_0: p = .5$ against the alternative $H_a: p > .5$ you need to
compute the test statistic
$$Z = \frac{\hat p - p_0}{\sqrt{p_0(1-p_0)/n}}.$$
In your example, you have $\hat p = 0.51,\,p_0 = 0.5,$ and $n = 1000.$
You would reject the null hypothesis $H_0,$ saying that the data from
the $n = 1000$ flips of the coin are inconsistent with the behavior
of a fair coin, at the 5% level if $Z > 1.645.$ (The 'critical value'
cutting 5% of the probability from the upper tail of a standard normal
distribution, can be found from printed normal CDF tables or from
software.)
I will leave it to you to finish this. Please compare this with what is
in your textbook, and leave a Comment if you still need help.
Notes: It would not be unusual to get 510 or more Heads in 1000 tosses of a fair
coin. That would happen over 25% of the time. A truly unusual outcome
would be to get exactly 500 Heads in 1000 tosses (probability about 0.025).
Results from R statistical software:
1 - pbinom(509, 1000, .5)
## 0.2739864 # P(510 Heads or more)
dbinom(500, 1000, .5)
## 0.02522502 $ P(EXACTLY 500 heads)