In that case the answer is "There is no way to tell."
You made 10 gal of wort with specific gravity of 1.096. Divide the 96 by 4.2 to get 23.3 °P. That's the percentage of sugar in your wort by weight. Now get the weight of the wort by multiplying its volume (10 gal) times the weight of a gallon, 8.35 times the specific gravity: 10*8.35*1.096 = 91.5 lbs. 23.3% of that is 0.233*10*8.35*1.096 = 21.32 pounds of extract. Divide that by the weight of the grain you mashed to get your efficiency. Now add the 1.25 lbs of sugar (sugar is 100% extract) to the 21.32 you got from the mash: 21.32 + 1.25 = 22.57. This is the effective extract in the fortified beer. The effective total weight of the wort also went up by 1.25 pounds so the effective original w/w percentage extract (OE) is
100*22.57/(91.5 + 1.25) = 24.32 °P. Multiply 24.32 by 4.2 to get 1.102 as the effective original gravity (OG). The only part that is at all tricky here is knowing the factor (4.2) which converts between SG and Plato. It is
fs = 3.8729 + 0.003276*points or, going in the other direction, fp = 3.8701 + 0.013796*°P. You can also do the conversion using the Lincoln equaitions, the ASBC polynomial, the ASBC or EBC tables.....
The fact that you added this at the end of fermentation and that it was, presumably, sucrose and must be inverted before fermentation implies that perhaps it will not be as completely fermented as the sugars from the mash so while the impact on OG is fairly easy to calculate the effect on alcohol content may not be.
