Answer:
The decision is to not reject the null hypothesis.
At a significance level of 0.01, there is not enough evidence to support the claim that the proportion of all those using the drug that experience relief is significantly higher than 50% (P-value = 0.1443).
Explanation:
This is a hypothesis test for a proportion.
The claim is that the proportion of all those using the drug that experience relief is significantly higher than 50%.
Then, the null and alternative hypothesis are:

The significance level is 0.01.
The sample has a size n=200.
The sample proportion is p=0.54.

The standard error of the proportion is:

Then, we can calculate the z-statistic as:

This test is a right-tailed test, so the P-value for this test is calculated as:

As the P-value (0.1443) is greater than the significance level (0.01), the effect is not significant.
The null hypothesis failed to be rejected.
At a significance level of 0.01, there is not enough evidence to support the claim that the proportion of all those using the drug that experience relief is significantly higher than 50%.