Cronbach's $\alpha$ is only designed for measures that are essentially $\tau$-equivalent, which essentially means they contribute equally to the underlying construct. One way to test this is to see if they have the same factor loadings in a factor model. If your measures are not $\tau$-equivalent, then $\alpha$ will underestimate reliability, regardless if the data are continuous or dichotomous.
There are many indices of internal-consistency reliability. $\omega$ (omega) as identified by McDonald (1999) is one of the most flexible for unidimensional constructs, and it can be easily extended to a multidimensional construct. Here is a procedure I recommend taking to identify which measure of reliability to use:
1) First assess dimensionality. Do you have 1 construct or many? If there are many, then no measure of unidimensional reliability will be accurate. Do this with factor analysis, ideally confirmatory factor analysis (CFA), but if you don't have the knowledge or the software you can use exploratory factor analysis (EFA). If you have more than 1 factor that is substantive, then you have a multidimensional construct. If that's the case, look for a measure of multidimensional reliability (these exist for both $\alpha$ and $\omega$. See here. Alternatively, identify the items that don't fit your desired construct and remove them (though take caution here, there are a lot of other psychometric tests you should do as well).
2) Assess $\tau$-equivalence. Again doing this in a factor model may be easiest. Basically, you test to see if the loadings are all equal - in a CFA, you can constrain the loadings and test fit, in an EFA you just have to ballpark the loadings to see if they are reasonably close. If you have $\tau$ equivalence, go ahead and use $\alpha$. If not, use $\omega$.
From what I can tell, SPSS does not calculate $\omega$ (see here). In my view, R is one of the best packages out there for psychometrics because it has the flexibility to do all of this. If you don't know R and don't have the time/energy to learn it (it's a big leap from SPSS) then you can probably safely go with $\alpha$ if you construct is unidimensional, just keep in mind reliability will be higher than what $\alpha$ gives you.
Reference:
McDonald, R. P. (1999). Test Theory: A Unified Treatment. New York: Psychology Press.
A confirmatory factor analysis would be used to test whether the questionnaire had the same structure among musicians as among athletes. EFA would be used to see what the factor structure is among musicians. Either might be right, but I think the latter is more likely what you want.
But before doing any FA, I'd look at item statistics. Perhaps some items that worked well with athletes do not work with musicians. Are there any items that get very skewed responses? They may not be contributing much. I would also look at all the correlations (39x38/2 of them). I might even make a huge scatterplot matrix to look at all the scatterplots.
Then I'd do EFA.
Best Answer
It seems that your intention is to measure 'perception of privacy' under each specific physical environment separately. In that case, your use of Cronbach's alpha would be to measure the internal consistency of items used to measure the 'perception of privacy' of your respondents in each environment. Hence calculate alpha for each physical environment and report your findings separately for each environment. You cannot average the alphas obtained from each scenario because that average doesn't make sense. It is also not possible to sum the ratings for all items across all possible environments to come up with a 'global perception of privacy' score because basically the scale you used is a unidimensional scale applied under different circumstances.
Also look at this post on CV.