Can this statistics question be solved as it stands? I have been presented with the following question:

An attorney claims that more than 25% of all lawyers advertise. A sample of 200 lawyers in a certain city showed that 63 had used some form of advertisement. At $\alpha$ = 0.05 is there enough evidence to support the attorney's claim? Use the P-value method. 

I don't know what method I should use to solve it, because it seems as though a standard deviation is needed in order to generate a test-statistic. I believe a z-score test will be used but I can't be positive. Can anyone shed light on this situation? 
Thanks!
 A: This is a simple binomial hypothesis test.
You should have the null hypothesis that the underlying probability ($\pi$) is exactly 25%.
(Sometimes this might be stated as less than or equal to 25%, but during the test we'll assume that this null hypothesis is true by using the 25% value.)
The alternative hypothesis is that the underlying probability is greater than 25% (matching the attorney's claim)
Now you want to work out the probability of the (right-hand) tail, based on the observed result. That is: P(X$\geq$63). This is called the p-value, as you require. Here, you work out the p-value according to a Binomial distribution matching the null hypothesis [X~B(200,0.25)].
The point of the p-value is to decide whether this result could reasonably be generated by the null hypothesis. 
If this resulting p-value is smaller or equal to (unlikelier) than $\alpha$=5%, then the result is "significant" - the result does not appear to be compatible with the null hypothesis, so therefore we'd accept the alternative hypothesis. (So, there is evidence the attorney's claim is correct)
If the resulting p-value is bigger than 5%, the result is "not significant" - the result appers to be compatible with the null hypothesis, so we accept the null hypothesis. (So, there is no evidence the attorney's claim is correct)
Edited To respond to your idea: a z-score would only apply to a test if we use a Normal distribution. Since this test is using a Binomial distribution, we don't really have any concept of a z-score here. 
This idea is quite subtle so feel free to ask questions if you're not sure.
A: Since $63/200 \gt 25\%$, any sensible approach should say the evidence strengthens the claim, though the evidence does not guarantee it.  
I suspect you are intended to do something like the following, though I dislike this approach myself.
Set the (one-sided) null hypothesis as "25% or fewer lawyers advertise": then the probability that as many as 63 or more out of 200 are found in the sample is 0.023 or less and so, since this is less than 0.05, you might reject the null hypothesis at this confidence level in favour of the claim.
