In order to answer this question, you need to specify how large a difference from $p_0$ you want to be 80% sure to detect.
Below is a printout from the 'power and sample size' procedure for one-sample tests of proportion in Minitab
statistical software. I assumed a test of $H_0: p = .5$ against
$H_a: p \ne .5,$ at the 0.05 = 5% level of significance with power .8 = 80%
for specific alternative values $p_a = .6, .7, .8.$
Power and Sample Size
Test for One Proportion
Testing p = 0.5 (versus ≠ 0.5)
α = 0.05
Sample Target
Comparison p Size Power Actual Power
0.6 194 0.8 0.800314
0.7 47 0.8 0.803325
0.8 20 0.8 0.817041
As you can see, if your null success value is $p_0 = .5,$ then it is easier to detect that the success probability is truly $p_a = 0.7$
than to detect $p_a = .6.$ So you need a smaller sample size for the former than for the latter. The following power curve from Minitab, shows
powers for various differences between $p_0 = .5$ and $p_a$ for each choice of sample size. Other statistical software programs (and with varying
degrees of clarity and accuracy, also some pages online) give similar output.
Notes: (1) In these computations, it is not just the distance
from the null value $p_0$ that matters, but also the null value itself.
For example, if $p_0 = .2,$ then the sample sizes necessary to detect
specific alternative values $p_a = .3, .4, .5$ are $n = 137, 36, 17,$ respectively. [Minitab output not shown.]
(2) With a sample size as large as $n = 194,$ a normal approximation to
the binomial distribution is good. So it would not matter whether
you use an exact binomial test or an approximate normal version of
the test. [You could find the critical values using standardization, solving quadratic equations in $n$ for closest integer values, and printed tables of the standard normal CDF. And then, more easily, you could verify the power.]
(3) Specifically, for testing $H_0: p = .5$ against $H_a: p > .5$ the
critical values for a test at level $\alpha = 0.5$ are $83$ and $111.$
That is $$P(X \le 83|p=.5)+P(X\ge 111|p=.5) \approx 0.05 = 5\%.$$
Also, such a test gives power $0.8 = 80\%$ against the alternative value $p_a = 0.6$ because
$$P(X \le 83|p=.6)+P(X\ge 111|p=.6) \approx 0.8 = 80\%.$$ [Because the binomial distribution is discrete, it is not possible for a nonrandomized test to have significance level exactly 5% or a power against a
specific alternative of exactly 80%.] Computations in R where dbinom is a binomial PDF:
sum(dbinom(c(0:83, 111:194), 194, .5))
[1] 0.05228312
sum(dbinom(c(0:83, 111:194), 194, .6))
[1] 0.8067075
In the figure below, the significance level of the test is the sum of
the heights of the blue bars in the rejection region outside the vertical dotted lines. The power of the test against the specific alternative
$p_a = 0.6$ is the sum of the heights of the brown bars in the rejection region---mainly in the the right tail. [Theoretically, both binomial distributions take values from 0 through 194, however probabilities of values of both distributions below 60 and above 140 are negligible and are not shown.]
R code for figure:
k =60:140; PDF = dbinom(k, 194, .5)
pdf.a = dbinom(k, 194, .6)
hdr = "PDFs of BINOM(194, .5) [blue] abd BINOM(194, .6)"
plot(k-.1, PDF, xlab="x", type="h", lwd=2, col="blue", main=hdr)
lines(k+.1, pdf.a, type="h", lwd=2, col="brown", main=hdr)
abline(h=0, col="green2")
abline(v=c(83.5,110.4), lty="dotted")
