Buffalo Bills: Josh Allen wins sixth Offensive Player of the Week award

Geoff Magliocchetti

Allen took home his fourth Player of the Week Award this season. Only Jim Kelly has earned more such titles in Buffalo Bills history.

Less than three seasons into his Buffalo Bills career, Josh Allen is already approaching Jim Kelly territory.

The NFL announced on Wednesday that the Bills’ quarterback has earned his fourth AFC Offensive Player of the Week Award of the season, stemming from a dominant performance in last Saturday’s visit to Denver. Allen put up 392 total yards of offense and four touchdowns (two each through the air and on the ground) in a 48-19 win over the Broncos, one that would allow the Bills to clinch their first AFC East title since 1995.

This season has been downright historic for Allen, whose emergence has been a large factor behind the Bills’ divisional takeover. Allen became the first quarterback in NFL history to score at least seven touchdowns in each of his first three seasons and he also joined Cam Newton and Daunte Culpepper as the only quarterbacks in league history to earn at least 60 scores through the air and at least 20 on the ground.

With the sixth Player of the Week award (with one each added in 2018 and 2019), Allen is now behind only Jim Kelly for the biggest haul in team history. This one for the Denver triumph broke a tie with Thurman Thomas with five and he also ties a single-season record for most Player of the Week Awards tallied by a Bill since O.J. Simpson in 1973.

Elsewhere in the AFC’s Week 15 affairs, Indianapolis lineman DeForest Buckner won the Defensive Player of the Week title, while Kansas City punter Tommy Townsend was the special teams representative.

Buffalo (11-3) returns to action on Monday night, taking on the New England Patriots in their final home game of the regular season (8:15 p.m. ET, ESPN/ABC).

Geoff Magliocchetti is on Twitter @GeoffJMags