I believe you have said previously the strings will get about 32V. Each LED draws 4V.
Conventional engineer's wisdom says to rate the resistor for twice what you expect it must carry.
For normal circumstances this would suggest using 1/2 W.
However, what if one LED wants to draw 3V instead of 4V?
The resistor must take on an extra 1 V.
At 150 mA current flow, that adds 150 mW to the normal 225 mW.
So a 1/2W resistor will not be overburdened. Flow through the string will increase a small amount, around 3 percent (or 1/32 of previous).
If an LED draws 2 V, however, then a 1/2 W resistor will be at its limit of endurance.
Current through the string will rise 6 percent.
Of course you would like to know how much variance to expect in the run of LED's.
In any case, it depends on which components you wish to protect.
Will an LED string continue working if one LED goes short-circuit? Then it's a question of: Will failure of one LED allow the product to continue being used? Or will it render it totally unusable?
If you want the product to continue working then you should increase the watt rating of the safety resistor. If not then you have to decide how to handle one LED failure, in that you don't want it to chain-reaction to other components.